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1671 lines
66 KiB
1671 lines
66 KiB
from fixtures import * # noqa: F401,F403
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from flaky import flaky
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from pyln.client import RpcError
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from utils import (
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only_one, sync_blockheight, wait_for, DEVELOPER, TIMEOUT, VALGRIND,
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SLOW_MACHINE, COMPAT
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)
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import os
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import queue
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import pytest
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import re
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import threading
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import unittest
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@unittest.skipIf(not DEVELOPER, "Too slow without --dev-bitcoind-poll")
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def test_closing(node_factory, bitcoind, chainparams):
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l1, l2 = node_factory.line_graph(2)
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chan = l1.get_channel_scid(l2)
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fee = 5430 if not chainparams['elements'] else 8955
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l1.pay(l2, 200000000)
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assert bitcoind.rpc.getmempoolinfo()['size'] == 0
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billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
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assert billboard == ['CHANNELD_NORMAL:Funding transaction locked.']
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billboard = only_one(l2.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status']
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assert billboard == ['CHANNELD_NORMAL:Funding transaction locked.']
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bitcoind.generate_block(5)
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# Only wait for the channels to activate with DEVELOPER=1,
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# otherwise it's going to take too long because of the missing
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# --dev-fast-gossip
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if DEVELOPER:
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wait_for(lambda: len(l1.getactivechannels()) == 2)
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wait_for(lambda: len(l2.getactivechannels()) == 2)
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billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
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# This may either be from a local_update or an announce, so just
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# check for the substring
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assert 'CHANNELD_NORMAL:Funding transaction locked.' in billboard[0]
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l1.rpc.close(chan)
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l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
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l2.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
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l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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# And should put closing into mempool.
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l1.daemon.wait_for_log('sendrawtx exit 0')
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l2.daemon.wait_for_log('sendrawtx exit 0')
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# Both nodes should have disabled the channel in their view
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wait_for(lambda: len(l1.getactivechannels()) == 0)
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wait_for(lambda: len(l2.getactivechannels()) == 0)
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assert bitcoind.rpc.getmempoolinfo()['size'] == 1
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# Now grab the close transaction
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closetxid = only_one(bitcoind.rpc.getrawmempool(False))
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billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
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assert billboard == [
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'CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of {} satoshi for tx:{}'.format(fee, closetxid),
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]
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bitcoind.generate_block(1)
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l1.daemon.wait_for_log(r'Owning output.* \(SEGWIT\).* txid %s.* CONFIRMED' % closetxid)
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l2.daemon.wait_for_log(r'Owning output.* \(SEGWIT\).* txid %s.* CONFIRMED' % closetxid)
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# Make sure both nodes have grabbed their close tx funds
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assert closetxid in set([o['txid'] for o in l1.rpc.listfunds()['outputs']])
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assert closetxid in set([o['txid'] for o in l2.rpc.listfunds()['outputs']])
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wait_for(lambda: only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status'] == [
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'CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of {} satoshi for tx:{}'.format(fee, closetxid),
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'ONCHAIN:Tracking mutual close transaction',
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'ONCHAIN:All outputs resolved: waiting 99 more blocks before forgetting channel'
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])
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bitcoind.generate_block(9)
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wait_for(lambda: only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status'] == [
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'CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of {} satoshi for tx:{}'.format(fee, closetxid),
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'ONCHAIN:Tracking mutual close transaction',
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'ONCHAIN:All outputs resolved: waiting 90 more blocks before forgetting channel'
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])
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# Make sure both have forgotten about it
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bitcoind.generate_block(90)
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wait_for(lambda: len(l1.rpc.listchannels()['channels']) == 0)
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wait_for(lambda: len(l2.rpc.listchannels()['channels']) == 0)
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# The entry in the channels table should still be there
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assert l1.db_query("SELECT count(*) as c FROM channels;")[0]['c'] == 1
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assert l2.db_query("SELECT count(*) as c FROM channels;")[0]['c'] == 1
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def test_closing_while_disconnected(node_factory, bitcoind, executor):
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l1, l2 = node_factory.line_graph(2, opts={'may_reconnect': True})
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chan = l1.get_channel_scid(l2)
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l1.pay(l2, 200000000)
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l2.stop()
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# The close should still be triggered afterwards.
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fut = executor.submit(l1.rpc.close, chan, 0)
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l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
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l2.start()
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fut.result(TIMEOUT)
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l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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# And should put closing into mempool.
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l1.daemon.wait_for_log('sendrawtx exit 0')
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l2.daemon.wait_for_log('sendrawtx exit 0')
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bitcoind.generate_block(101)
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wait_for(lambda: len(l1.rpc.listchannels()['channels']) == 0)
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wait_for(lambda: len(l2.rpc.listchannels()['channels']) == 0)
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def test_closing_id(node_factory):
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"""Test closing using peer ID and full channel ID
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"""
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l1, l2 = node_factory.get_nodes(2)
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# Close by full channel ID.
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l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
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l1.fund_channel(l2, 10**6)
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cid = l2.rpc.listpeers()['peers'][0]['channels'][0]['channel_id']
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l2.rpc.close(cid)
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wait_for(lambda: not only_one(l1.rpc.listpeers(l2.info['id'])['peers'])['connected'])
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wait_for(lambda: not only_one(l2.rpc.listpeers(l1.info['id'])['peers'])['connected'])
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# Close by peer ID.
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l2.rpc.connect(l1.info['id'], 'localhost', l1.port)
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l1.daemon.wait_for_log("Handed peer, entering loop")
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l2.fund_channel(l1, 10**6)
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pid = l1.info['id']
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l2.rpc.close(pid)
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wait_for(lambda: not only_one(l1.rpc.listpeers(l2.info['id'])['peers'])['connected'])
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wait_for(lambda: not only_one(l2.rpc.listpeers(l1.info['id'])['peers'])['connected'])
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@unittest.skipIf(VALGRIND, "Flaky under valgrind")
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def test_closing_torture(node_factory, executor, bitcoind):
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# We set up a fully-connected mesh of N nodes, then try
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# closing them all at once.
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amount = 10**6
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num_nodes = 10 # => 45 channels (36 seconds on my laptop)
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if VALGRIND:
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num_nodes -= 4 # => 15 (135 seconds)
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if SLOW_MACHINE:
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num_nodes -= 1 # => 36/10 (37/95 seconds)
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nodes = node_factory.get_nodes(num_nodes)
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# Make sure bitcoind has plenty of utxos
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bitcoind.generate_block(num_nodes)
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# Give them all plenty of UTXOs, make sure they see them
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for i in range(len(nodes)):
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for j in range(i + 1, len(nodes)):
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addr = nodes[i].rpc.newaddr()['bech32']
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bitcoind.rpc.sendtoaddress(addr, (amount + 1000000) / 10**8)
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bitcoind.generate_block(1)
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sync_blockheight(bitcoind, nodes)
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txs = []
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for i in range(len(nodes)):
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for j in range(i + 1, len(nodes)):
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nodes[i].rpc.connect(nodes[j].info['id'], 'localhost', nodes[j].port)
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txs.append(nodes[i].rpc.fundchannel(nodes[j].info['id'], amount)['txid'])
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# Make sure they're all in, then lock them in.
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bitcoind.generate_block(1, wait_for_mempool=txs)
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# Wait for them all to be CHANNELD_NORMAL
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for n in nodes:
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wait_for(lambda: all(p['channels'][0]['state'] == 'CHANNELD_NORMAL' for p in n.rpc.listpeers()['peers']))
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# Start closers: can take a long time under valgrind!
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futures = []
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for i in range(len(nodes)):
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for j in range(i + 1, len(nodes)):
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futures.append(executor.submit(nodes[i].rpc.close, nodes[j].info['id']))
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futures.append(executor.submit(nodes[j].rpc.close, nodes[i].info['id']))
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# Wait for close to finish
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close_txs = set()
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for f in futures:
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# If one side completes closing, we'll get an error here 'Peer has no active channel'
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try:
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close_txs.add(f.result(TIMEOUT)['txid'])
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except RpcError as err:
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assert err.error['message'] == 'Peer has no active channel'
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# Should have one close for each open.
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assert len(close_txs) == len(txs)
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# Get closes confirmed
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bitcoind.generate_block(100, wait_for_mempool=list(close_txs))
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# And make sure they hangup.
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for n in nodes:
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wait_for(lambda: n.rpc.listpeers()['peers'] == [])
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@unittest.skipIf(SLOW_MACHINE and VALGRIND, "slow test")
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def test_closing_different_fees(node_factory, bitcoind, executor):
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l1 = node_factory.get_node()
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# Default feerate = 15000/7500/1000
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# It will start at the second number, accepting anything above the first.
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feerates = [[20000, 15000, 7400], [8000, 1001, 100]]
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amounts = [0, 545999, 546000]
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num_peers = len(feerates) * len(amounts)
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addr = l1.rpc.newaddr()['bech32']
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bitcoind.rpc.sendtoaddress(addr, 1)
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numfunds = len(l1.rpc.listfunds()['outputs'])
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) > numfunds)
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# Create them in a batch, for speed!
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peers = []
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for feerate in feerates:
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for amount in amounts:
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p = node_factory.get_node(feerates=feerate)
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p.feerate = feerate
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p.amount = amount
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l1.rpc.connect(p.info['id'], 'localhost', p.port)
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peers.append(p)
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for p in peers:
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p.channel = l1.rpc.fundchannel(p.info['id'], 10**6, minconf=0)['channel_id']
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# Technically, this is async to fundchannel returning.
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l1.daemon.wait_for_log('sendrawtx exit 0')
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bitcoind.generate_block(6)
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# Now wait for them all to hit normal state, do payments
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l1.daemon.wait_for_logs(['update for channel .* now ACTIVE'] * num_peers
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+ ['to CHANNELD_NORMAL'] * num_peers)
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for p in peers:
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if p.amount != 0:
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l1.pay(p, 100000000)
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# Now close all channels (not unilaterally!)
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closes = [executor.submit(l1.rpc.close, p.channel, 0) for p in peers]
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for c in closes:
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c.result(90)
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# close does *not* wait for the sendrawtransaction, so do that!
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# Note that since they disagree on the ideal fee, they may conflict
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# (first one in will win), so we cannot look at logs, we need to
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# wait for mempool.
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wait_for(lambda: bitcoind.rpc.getmempoolinfo()['size'] == num_peers)
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bitcoind.generate_block(1)
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for p in peers:
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p.daemon.wait_for_log(' to ONCHAIN')
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wait_for(lambda: 'ONCHAIN:Tracking mutual close transaction' in only_one(p.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status'])
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l1.daemon.wait_for_logs([' to ONCHAIN'] * num_peers)
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@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
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def test_closing_negotiation_reconnect(node_factory, bitcoind):
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disconnects = ['-WIRE_CLOSING_SIGNED',
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'@WIRE_CLOSING_SIGNED',
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'+WIRE_CLOSING_SIGNED']
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l1 = node_factory.get_node(disconnect=disconnects, may_reconnect=True)
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l2 = node_factory.get_node(may_reconnect=True)
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l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
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chan = l1.fund_channel(l2, 10**6)
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l1.pay(l2, 200000000)
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assert bitcoind.rpc.getmempoolinfo()['size'] == 0
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l1.rpc.close(chan)
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l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
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l2.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
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l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
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# And should put closing into mempool (happens async, so
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# CLOSINGD_COMPLETE may come first).
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l1.daemon.wait_for_logs(['sendrawtx exit 0', ' to CLOSINGD_COMPLETE'])
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l2.daemon.wait_for_logs(['sendrawtx exit 0', ' to CLOSINGD_COMPLETE'])
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assert bitcoind.rpc.getmempoolinfo()['size'] == 1
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@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
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def test_closing_specified_destination(node_factory, bitcoind, chainparams):
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l1, l2, l3, l4 = node_factory.get_nodes(4)
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l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
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l1.rpc.connect(l3.info['id'], 'localhost', l3.port)
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l1.rpc.connect(l4.info['id'], 'localhost', l4.port)
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chan12 = l1.fund_channel(l2, 10**6)
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chan13 = l1.fund_channel(l3, 10**6)
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chan14 = l1.fund_channel(l4, 10**6)
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l1.pay(l2, 100000000)
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l1.pay(l3, 100000000)
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l1.pay(l4, 100000000)
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bitcoind.generate_block(5)
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addr = chainparams['example_addr']
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l1.rpc.close(chan12, None, addr)
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l1.rpc.call('close', {'id': chan13, 'destination': addr})
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l1.rpc.call('close', [chan14, None, addr])
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l1.daemon.wait_for_logs([' to CLOSINGD_SIGEXCHANGE'] * 3)
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# Both nodes should have disabled the channel in their view
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wait_for(lambda: len(l1.getactivechannels()) == 0)
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wait_for(lambda: bitcoind.rpc.getmempoolinfo()['size'] == 3)
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# Now grab the close transaction
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closetxs = {}
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for i, n in enumerate([l2, l3, l4]):
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billboard = only_one(l1.rpc.listpeers(n.info['id'])['peers'][0]['channels'])['status'][0]
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m = re.search(r'CLOSINGD_SIGEXCHANGE.* tx:([a-f0-9]{64})', billboard)
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closetxs[n] = m.group(1)
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bitcoind.generate_block(1)
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sync_blockheight(bitcoind, [l1, l2, l3, l4])
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# l1 can't spent the output to addr.
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for txid in closetxs.values():
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assert not l1.daemon.is_in_log(r'Owning output.* \(SEGWIT\).* txid {}.* CONFIRMED'.format(txid))
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# Check the txid has at least 1 confirmation
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for n, txid in closetxs.items():
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n.daemon.wait_for_log(r'Owning output.* \(SEGWIT\).* txid {}.* CONFIRMED'.format(txid))
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for n in [l2, l3, l4]:
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# Make sure both nodes have grabbed their close tx funds
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closetx = closetxs[n]
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outputs = n.rpc.listfunds()['outputs']
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assert closetx in set([o['txid'] for o in outputs])
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output_num2 = [o for o in outputs if o['txid'] == closetx][0]['output']
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output_num1 = 0 if output_num2 == 1 else 1
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# Check the another address is addr
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assert addr == bitcoind.rpc.gettxout(closetx, output_num1)['scriptPubKey']['addresses'][0]
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assert 1 == bitcoind.rpc.gettxout(closetx, output_num1)['confirmations']
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@unittest.skipIf(not COMPAT, "needs COMPAT=1")
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def test_deprecated_closing_compat(node_factory, bitcoind, chainparams):
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""" The old-style close command is:
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close {id} {force} {timeout}
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"""
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l1, l2 = node_factory.get_nodes(2, opts=[{'allow-deprecated-apis': True}, {}])
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addr = chainparams['example_addr']
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nodeid = l2.info['id']
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l1.rpc.check(command_to_check='close', id=nodeid)
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# New-style
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l1.rpc.check(command_to_check='close', id=nodeid, unilateraltimeout=10, destination=addr)
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l1.rpc.check(command_to_check='close', id=nodeid, unilateraltimeout=0)
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l1.rpc.check(command_to_check='close', id=nodeid, destination=addr)
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# Old-style
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l1.rpc.check(command_to_check='close', id=nodeid, force=False)
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l1.rpc.check(command_to_check='close', id=nodeid, force=False, timeout=10)
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l1.rpc.check(command_to_check='close', id=nodeid, timeout=10)
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l1.rpc.call('check', ['close', nodeid])
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# Array(new-style)
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l1.rpc.call('check', ['close', nodeid, 10])
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l1.rpc.call('check', ['close', nodeid, 0, addr])
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l1.rpc.call('check', ['close', nodeid, None, addr])
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# Array(old-style)
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l1.rpc.call('check', ['close', nodeid, True, 10])
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l1.rpc.call('check', ['close', nodeid, False])
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l1.rpc.call('check', ['close', nodeid, None, 10])
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# Not new-style nor old-style
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with pytest.raises(RpcError, match=r'Expected unilerataltimeout to be a number'):
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l1.rpc.call('check', ['close', nodeid, "Given enough eyeballs, all bugs are shallow."])
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@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
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def test_penalty_inhtlc(node_factory, bitcoind, executor, chainparams):
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"""Test penalty transaction with an incoming HTLC"""
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# We suppress each one after first commit; HTLC gets added not fulfilled.
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# Feerates identical so we don't get gratuitous commit to update them
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l1 = node_factory.get_node(disconnect=['=WIRE_COMMITMENT_SIGNED-nocommit'], may_fail=True, feerates=(7500, 7500, 7500), allow_broken_log=True)
|
|
l2 = node_factory.get_node(disconnect=['=WIRE_COMMITMENT_SIGNED-nocommit'])
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
# Now, this will get stuck due to l1 commit being disabled..
|
|
t = executor.submit(l1.pay, l2, 100000000)
|
|
|
|
assert len(l1.getactivechannels()) == 2
|
|
assert len(l2.getactivechannels()) == 2
|
|
|
|
# They should both have commitments blocked now.
|
|
l1.daemon.wait_for_log('=WIRE_COMMITMENT_SIGNED-nocommit')
|
|
l2.daemon.wait_for_log('=WIRE_COMMITMENT_SIGNED-nocommit')
|
|
|
|
# Make sure l1 got l2's commitment to the HTLC, and sent to master.
|
|
l1.daemon.wait_for_log('got commitsig')
|
|
|
|
# Take our snapshot.
|
|
tx = l1.rpc.dev_sign_last_tx(l2.info['id'])['tx']
|
|
|
|
# Let them continue
|
|
l1.rpc.dev_reenable_commit(l2.info['id'])
|
|
l2.rpc.dev_reenable_commit(l1.info['id'])
|
|
|
|
# Should fulfill.
|
|
l1.daemon.wait_for_log('peer_in WIRE_UPDATE_FULFILL_HTLC')
|
|
l1.daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
|
|
|
|
l2.daemon.wait_for_log('peer_out WIRE_UPDATE_FULFILL_HTLC')
|
|
l1.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
|
|
# Payment should now complete.
|
|
t.result(timeout=10)
|
|
|
|
# Now we really mess things up!
|
|
bitcoind.rpc.sendrawtransaction(tx)
|
|
bitcoind.generate_block(1)
|
|
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
# FIXME: l1 should try to stumble along!
|
|
wait_for(lambda: len(l2.getactivechannels()) == 0)
|
|
|
|
# l2 should spend all of the outputs (except to-us).
|
|
# Could happen in any order, depending on commitment tx.
|
|
needle = l2.daemon.logsearch_start
|
|
l2.wait_for_onchaind_broadcast('OUR_PENALTY_TX',
|
|
'THEIR_REVOKED_UNILATERAL/DELAYED_OUTPUT_TO_THEM')
|
|
l2.daemon.logsearch_start = needle
|
|
l2.wait_for_onchaind_broadcast('OUR_PENALTY_TX',
|
|
'THEIR_REVOKED_UNILATERAL/THEIR_HTLC')
|
|
|
|
# FIXME: test HTLC tx race!
|
|
|
|
# 100 blocks later, all resolved.
|
|
bitcoind.generate_block(100)
|
|
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
outputs = l2.rpc.listfunds()['outputs']
|
|
assert [o['status'] for o in outputs] == ['confirmed'] * 2
|
|
# Allow some lossage for fees.
|
|
slack = 27000 if chainparams['elements'] else 15000
|
|
assert sum(o['value'] for o in outputs) < 10**6
|
|
assert sum(o['value'] for o in outputs) > 10**6 - slack
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_penalty_outhtlc(node_factory, bitcoind, executor, chainparams):
|
|
"""Test penalty transaction with an outgoing HTLC"""
|
|
# First we need to get funds to l2, so suppress after second.
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(disconnect=['=WIRE_COMMITMENT_SIGNED*3-nocommit'], may_fail=True, feerates=(7500, 7500, 7500), allow_broken_log=True)
|
|
l2 = node_factory.get_node(disconnect=['=WIRE_COMMITMENT_SIGNED*3-nocommit'])
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
# Move some across to l2.
|
|
l1.pay(l2, 200000000)
|
|
|
|
assert not l1.daemon.is_in_log('=WIRE_COMMITMENT_SIGNED')
|
|
assert not l2.daemon.is_in_log('=WIRE_COMMITMENT_SIGNED')
|
|
|
|
# Now, this will get stuck due to l1 commit being disabled..
|
|
t = executor.submit(l2.pay, l1, 100000000)
|
|
|
|
# Make sure we get signature from them.
|
|
l1.daemon.wait_for_log('peer_in WIRE_UPDATE_ADD_HTLC')
|
|
l1.daemon.wait_for_log('peer_in WIRE_COMMITMENT_SIGNED')
|
|
|
|
# They should both have commitments blocked now.
|
|
l1.daemon.wait_for_log('dev_disconnect: =WIRE_COMMITMENT_SIGNED')
|
|
l2.daemon.wait_for_log('dev_disconnect: =WIRE_COMMITMENT_SIGNED')
|
|
|
|
# Make sure both sides got revoke_and_ack for that commitment.
|
|
l1.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
l2.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
|
|
# Take our snapshot.
|
|
tx = l1.rpc.dev_sign_last_tx(l2.info['id'])['tx']
|
|
|
|
# Let them continue
|
|
l1.rpc.dev_reenable_commit(l2.info['id'])
|
|
l2.rpc.dev_reenable_commit(l1.info['id'])
|
|
|
|
# Thread should complete.
|
|
t.result(timeout=10)
|
|
|
|
# Make sure both sides got revoke_and_ack for final.
|
|
l1.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
l2.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
|
|
# Now we really mess things up!
|
|
bitcoind.rpc.sendrawtransaction(tx)
|
|
bitcoind.generate_block(1)
|
|
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
# FIXME: l1 should try to stumble along!
|
|
|
|
# l2 should spend all of the outputs (except to-us).
|
|
# Could happen in any order, depending on commitment tx.
|
|
needle = l2.daemon.logsearch_start
|
|
l2.wait_for_onchaind_broadcast('OUR_PENALTY_TX',
|
|
'THEIR_REVOKED_UNILATERAL/DELAYED_OUTPUT_TO_THEM')
|
|
l2.daemon.logsearch_start = needle
|
|
l2.wait_for_onchaind_broadcast('OUR_PENALTY_TX',
|
|
'THEIR_REVOKED_UNILATERAL/OUR_HTLC')
|
|
|
|
l2.daemon.logsearch_start = needle
|
|
l2.daemon.wait_for_log('Ignoring output.*: THEIR_REVOKED_UNILATERAL/OUTPUT_TO_US')
|
|
|
|
# FIXME: test HTLC tx race!
|
|
|
|
# 100 blocks later, all resolved.
|
|
bitcoind.generate_block(100)
|
|
|
|
wait_for(lambda: len(l2.rpc.listpeers()['peers']) == 0)
|
|
|
|
outputs = l2.rpc.listfunds()['outputs']
|
|
assert [o['status'] for o in outputs] == ['confirmed'] * 3
|
|
# Allow some lossage for fees.
|
|
slack = 27000 if chainparams['elements'] else 15000
|
|
assert sum(o['value'] for o in outputs) < 10**6
|
|
assert sum(o['value'] for o in outputs) > 10**6 - slack
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_first_commit(node_factory, bitcoind):
|
|
"""Onchain handling where funder immediately drops to chain"""
|
|
|
|
# HTLC 1->2, 1 fails just after funding.
|
|
disconnects = ['+WIRE_FUNDING_LOCKED', 'permfail']
|
|
l1 = node_factory.get_node(disconnect=disconnects)
|
|
# Make locktime different, as we once had them reversed!
|
|
l2 = node_factory.get_node(options={'watchtime-blocks': 10})
|
|
l1.fundwallet(10**7)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
|
|
l1.rpc.fundchannel(l2.info['id'], 10**6)
|
|
l1.daemon.wait_for_log('sendrawtx exit 0')
|
|
|
|
l1.bitcoin.generate_block(1)
|
|
|
|
# l1 will drop to chain.
|
|
l1.daemon.wait_for_log('permfail')
|
|
l1.daemon.wait_for_log('sendrawtx exit 0')
|
|
l1.bitcoin.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# 10 later, l1 should collect its to-self payment.
|
|
bitcoind.generate_block(10)
|
|
l1.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# 94 later, l2 is done.
|
|
bitcoind.generate_block(94)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Now, 100 blocks and l1 should be done.
|
|
bitcoind.generate_block(6)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_unwatch(node_factory, bitcoind):
|
|
"""Onchaind should not watch random spends"""
|
|
l1, l2 = node_factory.line_graph(2)
|
|
|
|
l1.pay(l2, 200000000)
|
|
|
|
l1.rpc.dev_fail(l2.info['id'])
|
|
l1.daemon.wait_for_log('Failing due to dev-fail command')
|
|
l1.wait_for_channel_onchain(l2.info['id'])
|
|
|
|
l1.bitcoin.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# 10 later, l1 should collect its to-self payment.
|
|
bitcoind.generate_block(10)
|
|
l1.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# First time it sees it, onchaind cares.
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('Resolved OUR_UNILATERAL/DELAYED_OUTPUT_TO_US by our proposal '
|
|
'OUR_DELAYED_RETURN_TO_WALLET')
|
|
|
|
# Now test unrelated onchain churn.
|
|
# Daemon gets told about wallet; says it doesn't care.
|
|
l1.rpc.withdraw(l1.rpc.newaddr()['bech32'], 'all')
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log("but we don't care")
|
|
|
|
# And lightningd should respect that!
|
|
assert not l1.daemon.is_in_log("Can't unwatch txid")
|
|
|
|
# So these should not generate further messages
|
|
for i in range(5):
|
|
l1.rpc.withdraw(l1.rpc.newaddr()['bech32'], 'all')
|
|
bitcoind.generate_block(1)
|
|
# Make sure it digests the block
|
|
sync_blockheight(bitcoind, [l1])
|
|
|
|
# We won't see this again.
|
|
assert not l1.daemon.is_in_log("but we don't care",
|
|
start=l1.daemon.logsearch_start)
|
|
|
|
# Note: for this test we leave onchaind running, so we can detect
|
|
# any leaks!
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchaind_replay(node_factory, bitcoind):
|
|
disconnects = ['+WIRE_REVOKE_AND_ACK', 'permfail']
|
|
options = {'watchtime-blocks': 201, 'cltv-delta': 101}
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(options=options, disconnect=disconnects, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node(options=options)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
rhash = l2.rpc.invoice(10**8, 'onchaind_replay', 'desc')['payment_hash']
|
|
routestep = {
|
|
'msatoshi': 10**8 - 1,
|
|
'id': l2.info['id'],
|
|
'delay': 101,
|
|
'channel': '1x1x1'
|
|
}
|
|
l1.rpc.sendpay([routestep], rhash)
|
|
l1.daemon.wait_for_log('sendrawtx exit 0')
|
|
bitcoind.generate_block(1)
|
|
|
|
# Wait for nodes to notice the failure, this seach needle is after the
|
|
# DB commit so we're sure the tx entries in onchaindtxs have been added
|
|
l1.daemon.wait_for_log("Deleting channel .* due to the funding outpoint being spent")
|
|
l2.daemon.wait_for_log("Deleting channel .* due to the funding outpoint being spent")
|
|
|
|
# We should at least have the init tx now
|
|
assert len(l1.db_query("SELECT * FROM channeltxs;")) > 0
|
|
assert len(l2.db_query("SELECT * FROM channeltxs;")) > 0
|
|
|
|
# Generate some blocks so we restart the onchaind from DB (we rescan
|
|
# last_height - 100)
|
|
bitcoind.generate_block(100)
|
|
sync_blockheight(bitcoind, [l1, l2])
|
|
|
|
# l1 should still have a running onchaind
|
|
assert len(l1.db_query("SELECT * FROM channeltxs;")) > 0
|
|
|
|
l2.rpc.stop()
|
|
l1.restart()
|
|
|
|
# Can't wait for it, it's after the "Server started" wait in restart()
|
|
assert l1.daemon.is_in_log(r'Restarting onchaind for channel')
|
|
|
|
# l1 should still notice that the funding was spent and that we should react to it
|
|
l1.daemon.wait_for_log("Propose handling OUR_UNILATERAL/DELAYED_OUTPUT_TO_US by OUR_DELAYED_RETURN_TO_WALLET")
|
|
sync_blockheight(bitcoind, [l1])
|
|
bitcoind.generate_block(10)
|
|
sync_blockheight(bitcoind, [l1])
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_dust_out(node_factory, bitcoind, executor):
|
|
"""Onchain handling of outgoing dust htlcs (they should fail)"""
|
|
# HTLC 1->2, 1 fails after it's irrevocably committed
|
|
disconnects = ['@WIRE_REVOKE_AND_ACK', 'permfail']
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(disconnect=disconnects, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node()
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
# Must be dust!
|
|
rhash = l2.rpc.invoice(1, 'onchain_dust_out', 'desc')['payment_hash']
|
|
routestep = {
|
|
'msatoshi': 1,
|
|
'id': l2.info['id'],
|
|
'delay': 5,
|
|
'channel': '1x1x1'
|
|
}
|
|
|
|
l1.rpc.sendpay([routestep], rhash)
|
|
payfuture = executor.submit(l1.rpc.waitsendpay, rhash)
|
|
|
|
# l1 will drop to chain.
|
|
l1.daemon.wait_for_log('permfail')
|
|
l1.wait_for_channel_onchain(l2.info['id'])
|
|
l1.bitcoin.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# We use 3 blocks for "reasonable depth"
|
|
bitcoind.generate_block(3)
|
|
|
|
# It should fail.
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE: missing in commitment tx'):
|
|
payfuture.result(5)
|
|
|
|
# Retry payment, this should fail (and, as a side-effect, tickle a
|
|
# bug).
|
|
with pytest.raises(RpcError, match=r'WIRE_UNKNOWN_NEXT_PEER'):
|
|
l1.rpc.sendpay([routestep], rhash)
|
|
|
|
# 6 later, l1 should collect its to-self payment.
|
|
bitcoind.generate_block(6)
|
|
l1.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# 94 later, l2 is done.
|
|
bitcoind.generate_block(94)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Restart l1, it should not crash!
|
|
l1.restart()
|
|
|
|
# Now, 100 blocks and l1 should be done.
|
|
bitcoind.generate_block(6)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Payment failed, BTW
|
|
assert only_one(l2.rpc.listinvoices('onchain_dust_out')['invoices'])['status'] == 'unpaid'
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_timeout(node_factory, bitcoind, executor):
|
|
"""Onchain handling of outgoing failed htlcs"""
|
|
# HTLC 1->2, 1 fails just after it's irrevocably committed
|
|
disconnects = ['+WIRE_REVOKE_AND_ACK*3', 'permfail']
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(disconnect=disconnects, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node()
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
rhash = l2.rpc.invoice(10**8, 'onchain_timeout', 'desc')['payment_hash']
|
|
# We underpay, so it fails.
|
|
routestep = {
|
|
'msatoshi': 10**8 - 1,
|
|
'id': l2.info['id'],
|
|
'delay': 5,
|
|
'channel': '1x1x1'
|
|
}
|
|
|
|
l1.rpc.sendpay([routestep], rhash)
|
|
with pytest.raises(RpcError):
|
|
l1.rpc.waitsendpay(rhash)
|
|
|
|
# Make sure CLTVs are different, in case it confuses onchaind.
|
|
bitcoind.generate_block(1)
|
|
sync_blockheight(bitcoind, [l1])
|
|
|
|
# Second one will cause drop to chain.
|
|
l1.rpc.sendpay([routestep], rhash)
|
|
payfuture = executor.submit(l1.rpc.waitsendpay, rhash)
|
|
|
|
# l1 will drop to chain.
|
|
l1.daemon.wait_for_log('permfail')
|
|
l1.wait_for_channel_onchain(l2.info['id'])
|
|
l1.bitcoin.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Wait for timeout.
|
|
l1.daemon.wait_for_logs(['Propose handling OUR_UNILATERAL/DELAYED_OUTPUT_TO_US by OUR_DELAYED_RETURN_TO_WALLET .* after 5 blocks',
|
|
'Propose handling OUR_UNILATERAL/OUR_HTLC by OUR_HTLC_TIMEOUT_TX .* after 6 blocks'])
|
|
bitcoind.generate_block(4)
|
|
|
|
l1.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
bitcoind.generate_block(1)
|
|
l1.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
|
|
# We use 3 blocks for "reasonable depth"
|
|
bitcoind.generate_block(3)
|
|
|
|
# It should fail.
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE: timed out'):
|
|
payfuture.result(5)
|
|
|
|
# 2 later, l1 spends HTLC (5 blocks total).
|
|
bitcoind.generate_block(2)
|
|
l1.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
|
|
|
|
# 89 later, l2 is done.
|
|
bitcoind.generate_block(89)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Now, 100 blocks and l1 should be done.
|
|
bitcoind.generate_block(10)
|
|
sync_blockheight(bitcoind, [l1])
|
|
assert not l1.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Payment failed, BTW
|
|
assert only_one(l2.rpc.listinvoices('onchain_timeout')['invoices'])['status'] == 'unpaid'
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_middleman(node_factory, bitcoind):
|
|
# HTLC 1->2->3, 1->2 goes down after 2 gets preimage from 3.
|
|
disconnects = ['-WIRE_UPDATE_FULFILL_HTLC', 'permfail']
|
|
l1 = node_factory.get_node()
|
|
l2 = node_factory.get_node(disconnect=disconnects)
|
|
l3 = node_factory.get_node()
|
|
|
|
# l2 connects to both, so l1 can't reconnect and thus l2 drops to chain
|
|
l2.rpc.connect(l1.info['id'], 'localhost', l1.port)
|
|
l2.rpc.connect(l3.info['id'], 'localhost', l3.port)
|
|
l2.fund_channel(l1, 10**6)
|
|
c23 = l2.fund_channel(l3, 10**6)
|
|
|
|
# Make sure routes finalized.
|
|
bitcoind.generate_block(5)
|
|
l1.wait_channel_active(c23)
|
|
|
|
# Give l1 some money to play with.
|
|
l2.pay(l1, 2 * 10**8)
|
|
|
|
# Must be bigger than dust!
|
|
rhash = l3.rpc.invoice(10**8, 'middleman', 'desc')['payment_hash']
|
|
|
|
route = l1.rpc.getroute(l3.info['id'], 10**8, 1)["route"]
|
|
assert len(route) == 2
|
|
|
|
q = queue.Queue()
|
|
|
|
def try_pay():
|
|
try:
|
|
l1.rpc.sendpay(route, rhash)
|
|
l1.rpc.waitsendpay(rhash)
|
|
q.put(None)
|
|
except Exception as err:
|
|
q.put(err)
|
|
|
|
t = threading.Thread(target=try_pay)
|
|
t.daemon = True
|
|
t.start()
|
|
|
|
# l2 will drop to chain.
|
|
l2.daemon.wait_for_log('sendrawtx exit 0')
|
|
l1.bitcoin.generate_block(1)
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log('OUR_UNILATERAL/THEIR_HTLC')
|
|
|
|
# l2 should fulfill HTLC onchain, and spend to-us (any order)
|
|
l2.wait_for_onchaind_broadcast('OUR_HTLC_SUCCESS_TX',
|
|
'OUR_UNILATERAL/THEIR_HTLC')
|
|
|
|
# Payment should succeed.
|
|
l1.bitcoin.generate_block(1)
|
|
l1.daemon.wait_for_log('THEIR_UNILATERAL/OUR_HTLC gave us preimage')
|
|
err = q.get(timeout=10)
|
|
if err:
|
|
print("Got err from sendpay thread")
|
|
raise err
|
|
t.join(timeout=1)
|
|
assert not t.is_alive()
|
|
|
|
# Three more, l2 can spend to-us.
|
|
bitcoind.generate_block(3)
|
|
l2.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# One more block, HTLC tx is now spendable.
|
|
l1.bitcoin.generate_block(1)
|
|
l2.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_HTLC_SUCCESS_TX/DELAYED_OUTPUT_TO_US')
|
|
|
|
# 100 blocks after last spend, l2 should be done.
|
|
l1.bitcoin.generate_block(100)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_onchain_feechange(node_factory, bitcoind, executor):
|
|
"""Onchain handling when we restart with different fees"""
|
|
# HTLC 1->2, 2 fails just after they're both irrevocably committed
|
|
# We need 2 to drop to chain, because then 1's HTLC timeout tx
|
|
# is generated on-the-fly, and is thus feerate sensitive.
|
|
disconnects = ['-WIRE_UPDATE_FAIL_HTLC', 'permfail']
|
|
l1 = node_factory.get_node(may_reconnect=True)
|
|
l2 = node_factory.get_node(disconnect=disconnects,
|
|
may_reconnect=True)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
rhash = l2.rpc.invoice(10**8, 'onchain_timeout', 'desc')['payment_hash']
|
|
# We underpay, so it fails.
|
|
routestep = {
|
|
'msatoshi': 10**8 - 1,
|
|
'id': l2.info['id'],
|
|
'delay': 5,
|
|
'channel': '1x1x1'
|
|
}
|
|
|
|
executor.submit(l1.rpc.sendpay, [routestep], rhash)
|
|
|
|
# l2 will drop to chain.
|
|
l2.daemon.wait_for_log('permfail')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Wait for timeout.
|
|
l1.daemon.wait_for_log('Propose handling THEIR_UNILATERAL/OUR_HTLC by OUR_HTLC_TIMEOUT_TO_US .* after 6 blocks')
|
|
bitcoind.generate_block(6)
|
|
|
|
l1.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
|
|
# Make sure that gets included.
|
|
|
|
bitcoind.generate_block(1)
|
|
# Now we restart with different feerates.
|
|
l1.stop()
|
|
|
|
l1.daemon.cmd_line.append('--override-fee-rates=20000/9000/2000')
|
|
l1.start()
|
|
|
|
# We recognize different proposal as ours.
|
|
l1.daemon.wait_for_log('Resolved THEIR_UNILATERAL/OUR_HTLC by our proposal OUR_HTLC_TIMEOUT_TO_US')
|
|
|
|
# We use 3 blocks for "reasonable depth", so add two more
|
|
bitcoind.generate_block(2)
|
|
|
|
# Note that the very similar test_onchain_timeout looks for a
|
|
# different string: that's because it sees the JSONRPC response,
|
|
# and due to the l1 restart, there is none here.
|
|
l1.daemon.wait_for_log('WIRE_PERMANENT_CHANNEL_FAILURE')
|
|
|
|
# 90 later, l2 is done
|
|
bitcoind.generate_block(89)
|
|
sync_blockheight(bitcoind, [l2])
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Now, 7 blocks and l1 should be done.
|
|
bitcoind.generate_block(6)
|
|
sync_blockheight(bitcoind, [l1])
|
|
assert not l1.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# Payment failed, BTW
|
|
assert only_one(l2.rpc.listinvoices('onchain_timeout')['invoices'])['status'] == 'unpaid'
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1 for dev-set-fees")
|
|
def test_onchain_all_dust(node_factory, bitcoind, executor):
|
|
"""Onchain handling when we reduce output to all dust"""
|
|
# HTLC 1->2, 2 fails just after they're both irrevocably committed
|
|
# We need 2 to drop to chain, because then 1's HTLC timeout tx
|
|
# is generated on-the-fly, and is thus feerate sensitive.
|
|
disconnects = ['-WIRE_UPDATE_FAIL_HTLC', 'permfail']
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(options={'dev-no-reconnect': None}, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node(disconnect=disconnects)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
rhash = l2.rpc.invoice(10**8, 'onchain_timeout', 'desc')['payment_hash']
|
|
# We underpay, so it fails.
|
|
routestep = {
|
|
'msatoshi': 10**7 - 1,
|
|
'id': l2.info['id'],
|
|
'delay': 5,
|
|
'channel': '1x1x1'
|
|
}
|
|
|
|
executor.submit(l1.rpc.sendpay, [routestep], rhash)
|
|
|
|
# l2 will drop to chain.
|
|
l2.daemon.wait_for_log('permfail')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
|
|
# Make l1's fees really high (and wait for it to exceed 50000)
|
|
l1.set_feerates((100000, 100000, 100000))
|
|
l1.daemon.wait_for_log('Feerate estimate for normal set to [56789][0-9]{4}')
|
|
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Wait for timeout.
|
|
l1.daemon.wait_for_log('Propose handling THEIR_UNILATERAL/OUR_HTLC by IGNORING_TINY_PAYMENT .* after 6 blocks')
|
|
bitcoind.generate_block(5)
|
|
|
|
l1.wait_for_onchaind_broadcast('IGNORING_TINY_PAYMENT',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
l1.daemon.wait_for_log('Ignoring output 0 of .*: THEIR_UNILATERAL/OUR_HTLC')
|
|
|
|
# 100 deep and l2 forgets.
|
|
bitcoind.generate_block(93)
|
|
sync_blockheight(bitcoind, [l1, l2])
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
assert not l1.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
# l1 does not wait for ignored payment.
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1 for dev_fail")
|
|
def test_onchain_different_fees(node_factory, bitcoind, executor):
|
|
"""Onchain handling when we've had a range of fees"""
|
|
l1, l2 = node_factory.line_graph(2, fundchannel=True, fundamount=10**7,
|
|
opts={'may_reconnect': True})
|
|
|
|
l2.rpc.dev_ignore_htlcs(id=l1.info['id'], ignore=True)
|
|
p1 = executor.submit(l1.pay, l2, 1000000000)
|
|
l1.daemon.wait_for_log('htlc 0: RCVD_ADD_ACK_COMMIT->SENT_ADD_ACK_REVOCATION')
|
|
|
|
l1.set_feerates((16000, 7500, 3750))
|
|
p2 = executor.submit(l1.pay, l2, 900000000)
|
|
l1.daemon.wait_for_log('htlc 1: RCVD_ADD_ACK_COMMIT->SENT_ADD_ACK_REVOCATION')
|
|
|
|
# Restart with different feerate for second HTLC.
|
|
l1.set_feerates((5000, 5000, 3750))
|
|
l1.restart()
|
|
l1.daemon.wait_for_log('peer_out WIRE_UPDATE_FEE')
|
|
|
|
p3 = executor.submit(l1.pay, l2, 800000000)
|
|
l1.daemon.wait_for_log('htlc 2: RCVD_ADD_ACK_COMMIT->SENT_ADD_ACK_REVOCATION')
|
|
|
|
# Drop to chain
|
|
l1.rpc.dev_fail(l2.info['id'])
|
|
l1.wait_for_channel_onchain(l2.info['id'])
|
|
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Both sides should have correct feerate
|
|
assert l1.db_query('SELECT min_possible_feerate, max_possible_feerate FROM channels;') == [{
|
|
'min_possible_feerate': 5000,
|
|
'max_possible_feerate': 16000
|
|
}]
|
|
assert l2.db_query('SELECT min_possible_feerate, max_possible_feerate FROM channels;') == [{
|
|
'min_possible_feerate': 5000,
|
|
'max_possible_feerate': 16000
|
|
}]
|
|
|
|
bitcoind.generate_block(5)
|
|
# Three HTLCs, and one for the to-us output.
|
|
l1.daemon.wait_for_logs(['sendrawtx exit 0'] * 4)
|
|
|
|
# We use 3 blocks for "reasonable depth"
|
|
bitcoind.generate_block(3)
|
|
|
|
with pytest.raises(Exception):
|
|
p1.result(10)
|
|
with pytest.raises(Exception):
|
|
p2.result(10)
|
|
with pytest.raises(Exception):
|
|
p3.result(10)
|
|
|
|
# Two more for HTLC timeout tx to be spent.
|
|
bitcoind.generate_block(2)
|
|
l1.daemon.wait_for_logs(['sendrawtx exit 0'] * 3)
|
|
|
|
# Now, 100 blocks it should be done.
|
|
bitcoind.generate_block(100)
|
|
wait_for(lambda: l1.rpc.listpeers()['peers'] == [])
|
|
wait_for(lambda: l2.rpc.listpeers()['peers'] == [])
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_permfail_new_commit(node_factory, bitcoind, executor):
|
|
# Test case where we have two possible commits: it will use new one.
|
|
disconnects = ['-WIRE_REVOKE_AND_ACK', 'permfail']
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(options={'dev-no-reconnect': None}, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node(disconnect=disconnects)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
# This will fail at l2's end.
|
|
t = executor.submit(l1.pay, l2, 200000000)
|
|
|
|
l2.daemon.wait_for_log('dev_disconnect permfail')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('Their unilateral tx, new commit point')
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log('Propose handling OUR_UNILATERAL/THEIR_HTLC by THEIR_HTLC_TIMEOUT_TO_THEM \\(IGNORING\\) after 6 blocks')
|
|
l1.daemon.wait_for_log('Propose handling THEIR_UNILATERAL/OUR_HTLC by OUR_HTLC_TIMEOUT_TO_US (.*) after 6 blocks')
|
|
|
|
# OK, time out HTLC.
|
|
bitcoind.generate_block(5)
|
|
l1.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('Resolved THEIR_UNILATERAL/OUR_HTLC by our proposal OUR_HTLC_TIMEOUT_TO_US')
|
|
l2.daemon.wait_for_log('Ignoring output.*: OUR_UNILATERAL/THEIR_HTLC')
|
|
|
|
t.cancel()
|
|
|
|
# Now, 100 blocks it should be done.
|
|
bitcoind.generate_block(100)
|
|
wait_for(lambda: l1.rpc.listpeers()['peers'] == [])
|
|
wait_for(lambda: l2.rpc.listpeers()['peers'] == [])
|
|
|
|
|
|
def setup_multihtlc_test(node_factory, bitcoind):
|
|
# l1 -> l2 -> l3 -> l4 -> l5 -> l6 -> l7
|
|
# l1 and l7 ignore and HTLCs they're sent.
|
|
# For each direction, we create these HTLCs with same payment_hash:
|
|
# 1 failed (CLTV1)
|
|
# 1 failed (CLTV2)
|
|
# 2 live (CLTV2)
|
|
# 1 live (CLTV3)
|
|
nodes = node_factory.line_graph(7, wait_for_announce=True,
|
|
opts={'dev-no-reconnect': None,
|
|
'may_reconnect': True})
|
|
|
|
# Balance by pushing half the funds.
|
|
b11 = nodes[-1].rpc.invoice(10**9 // 2, '1', 'balancer')['bolt11']
|
|
nodes[0].rpc.pay(b11)
|
|
|
|
nodes[0].rpc.dev_ignore_htlcs(id=nodes[1].info['id'], ignore=True)
|
|
nodes[-1].rpc.dev_ignore_htlcs(id=nodes[-2].info['id'], ignore=True)
|
|
|
|
preimage = "0" * 64
|
|
h = nodes[0].rpc.invoice(msatoshi=10**8, label='x', description='desc',
|
|
preimage=preimage)['payment_hash']
|
|
nodes[-1].rpc.invoice(msatoshi=10**8, label='x', description='desc',
|
|
preimage=preimage)['payment_hash']
|
|
|
|
# First, the failed attempts (paying wrong node). CLTV1
|
|
r = nodes[0].rpc.getroute(nodes[-2].info['id'], 10**8, 1)["route"]
|
|
nodes[0].rpc.sendpay(r, h)
|
|
with pytest.raises(RpcError, match=r'INCORRECT_OR_UNKNOWN_PAYMENT_DETAILS'):
|
|
nodes[0].rpc.waitsendpay(h)
|
|
|
|
r = nodes[-1].rpc.getroute(nodes[1].info['id'], 10**8, 1)["route"]
|
|
nodes[-1].rpc.sendpay(r, h)
|
|
with pytest.raises(RpcError, match=r'INCORRECT_OR_UNKNOWN_PAYMENT_DETAILS'):
|
|
nodes[-1].rpc.waitsendpay(h)
|
|
|
|
# Now increment CLTV -> CLTV2
|
|
bitcoind.generate_block(1)
|
|
sync_blockheight(bitcoind, nodes)
|
|
|
|
# Now, the live attempts with CLTV2 (blackholed by end nodes)
|
|
r = nodes[0].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
|
|
nodes[0].rpc.sendpay(r, h)
|
|
r = nodes[-1].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
|
|
nodes[-1].rpc.sendpay(r, h)
|
|
|
|
# We send second HTLC from different node, since they refuse to send
|
|
# multiple with same hash.
|
|
r = nodes[1].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
|
|
nodes[1].rpc.sendpay(r, h)
|
|
r = nodes[-2].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
|
|
nodes[-2].rpc.sendpay(r, h)
|
|
|
|
# Now increment CLTV -> CLTV3.
|
|
bitcoind.generate_block(1)
|
|
sync_blockheight(bitcoind, nodes)
|
|
|
|
r = nodes[2].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
|
|
nodes[2].rpc.sendpay(r, h)
|
|
r = nodes[-3].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
|
|
nodes[-3].rpc.sendpay(r, h)
|
|
|
|
# Make sure HTLCs have reached the end.
|
|
nodes[0].daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 3)
|
|
nodes[-1].daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 3)
|
|
|
|
return h, nodes
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1 for dev_ignore_htlcs")
|
|
@unittest.skipIf(SLOW_MACHINE and VALGRIND, "slow test")
|
|
def test_onchain_multihtlc_our_unilateral(node_factory, bitcoind):
|
|
"""Node pushes a channel onchain with multiple HTLCs with same payment_hash """
|
|
h, nodes = setup_multihtlc_test(node_factory, bitcoind)
|
|
|
|
mid = len(nodes) // 2
|
|
|
|
for i in range(len(nodes) - 1):
|
|
assert only_one(nodes[i].rpc.listpeers(nodes[i + 1].info['id'])['peers'])['connected']
|
|
|
|
# Now midnode goes onchain with n+1 channel.
|
|
nodes[mid].rpc.dev_fail(nodes[mid + 1].info['id'])
|
|
nodes[mid].wait_for_channel_onchain(nodes[mid + 1].info['id'])
|
|
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].daemon.wait_for_log(' to ONCHAIN')
|
|
nodes[mid + 1].daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Now, restart and manually reconnect end nodes (so they don't ignore HTLCs)
|
|
# In fact, they'll fail them with WIRE_TEMPORARY_NODE_FAILURE.
|
|
# TODO Remove our reliance on HTLCs failing on startup and the need for
|
|
# this plugin
|
|
nodes[0].daemon.opts['plugin'] = os.path.join(os.getcwd(), 'tests/plugins/fail_htlcs.py')
|
|
nodes[-1].daemon.opts['plugin'] = os.path.join(os.getcwd(), 'tests/plugins/fail_htlcs.py')
|
|
nodes[0].restart()
|
|
nodes[-1].restart()
|
|
|
|
# We disabled auto-reconnect so we'd detect breakage, so manually reconnect.
|
|
nodes[0].rpc.connect(nodes[1].info['id'], 'localhost', nodes[1].port)
|
|
nodes[-1].rpc.connect(nodes[-2].info['id'], 'localhost', nodes[-2].port)
|
|
|
|
# Wait for HTLCs to stabilize.
|
|
nodes[0].daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
|
|
nodes[0].daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
|
|
nodes[0].daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
|
|
nodes[-1].daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
|
|
nodes[-1].daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
|
|
nodes[-1].daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
|
|
|
|
# After at depth 5, midnode will spend its own to-self output.
|
|
bitcoind.generate_block(4)
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# The three outgoing HTLCs time out at 21, 21 and 22 blocks.
|
|
bitcoind.generate_block(16)
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
|
|
# And three more for us to consider them all settled.
|
|
bitcoind.generate_block(3)
|
|
|
|
# Now, those nodes should have correctly failed the HTLCs
|
|
for n in nodes[:mid - 1]:
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
|
|
n.rpc.waitsendpay(h, TIMEOUT)
|
|
|
|
# Other timeouts are 27,27,28 blocks.
|
|
bitcoind.generate_block(2)
|
|
nodes[mid].daemon.wait_for_logs(['Ignoring output.*: OUR_UNILATERAL/THEIR_HTLC'] * 2)
|
|
for _ in range(2):
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].daemon.wait_for_log('Ignoring output.*: OUR_UNILATERAL/THEIR_HTLC')
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
|
|
# Depth 3 to consider it settled.
|
|
bitcoind.generate_block(3)
|
|
|
|
for n in nodes[mid + 1:]:
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
|
|
n.rpc.waitsendpay(h, TIMEOUT)
|
|
|
|
# At depth 100 it's all done (we didn't bother waiting for mid+1's
|
|
# spends, so that might still be going)
|
|
bitcoind.generate_block(97)
|
|
nodes[mid].daemon.wait_for_logs(['onchaind complete, forgetting peer'])
|
|
|
|
# No other channels should have failed.
|
|
for i in range(len(nodes) - 1):
|
|
if i != mid:
|
|
assert only_one(nodes[i].rpc.listpeers(nodes[i + 1].info['id'])['peers'])['connected']
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1 for dev_ignore_htlcs")
|
|
@unittest.skipIf(SLOW_MACHINE and VALGRIND, "slow test")
|
|
def test_onchain_multihtlc_their_unilateral(node_factory, bitcoind):
|
|
"""Node pushes a channel onchain with multiple HTLCs with same payment_hash """
|
|
h, nodes = setup_multihtlc_test(node_factory, bitcoind)
|
|
|
|
mid = len(nodes) // 2
|
|
|
|
for i in range(len(nodes) - 1):
|
|
assert only_one(nodes[i].rpc.listpeers(nodes[i + 1].info['id'])['peers'])['connected']
|
|
|
|
# Now midnode+1 goes onchain with midnode channel.
|
|
nodes[mid + 1].rpc.dev_fail(nodes[mid].info['id'])
|
|
nodes[mid + 1].wait_for_channel_onchain(nodes[mid].info['id'])
|
|
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].daemon.wait_for_log(' to ONCHAIN')
|
|
nodes[mid + 1].daemon.wait_for_log(' to ONCHAIN')
|
|
|
|
# Now, restart and manually reconnect end nodes (so they don't ignore HTLCs)
|
|
# In fact, they'll fail them with WIRE_TEMPORARY_NODE_FAILURE.
|
|
# TODO Remove our reliance on HTLCs failing on startup and the need for
|
|
# this plugin
|
|
nodes[0].daemon.opts['plugin'] = os.path.join(os.getcwd(), 'tests/plugins/fail_htlcs.py')
|
|
nodes[-1].daemon.opts['plugin'] = os.path.join(os.getcwd(), 'tests/plugins/fail_htlcs.py')
|
|
nodes[0].restart()
|
|
nodes[-1].restart()
|
|
|
|
# We disabled auto-reconnect so we'd detect breakage, so manually reconnect.
|
|
nodes[0].rpc.connect(nodes[1].info['id'], 'localhost', nodes[1].port)
|
|
nodes[-1].rpc.connect(nodes[-2].info['id'], 'localhost', nodes[-2].port)
|
|
|
|
# Wait for HTLCs to stabilize.
|
|
nodes[0].daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
|
|
nodes[0].daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
|
|
nodes[0].daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
|
|
nodes[-1].daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
|
|
nodes[-1].daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
|
|
nodes[-1].daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
|
|
|
|
# At depth 5, midnode+1 will spend its own to-self output.
|
|
bitcoind.generate_block(4)
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET')
|
|
|
|
# The three outgoing HTLCs time out at depth 21, 21 and 22 blocks.
|
|
bitcoind.generate_block(16)
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
|
|
'THEIR_UNILATERAL/OUR_HTLC')
|
|
|
|
# At depth 3 we consider them all settled.
|
|
bitcoind.generate_block(3)
|
|
|
|
# Now, those nodes should have correctly failed the HTLCs
|
|
for n in nodes[:mid - 1]:
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
|
|
n.rpc.waitsendpay(h, TIMEOUT)
|
|
|
|
# Other timeouts are at depths 27,27,28 blocks.
|
|
bitcoind.generate_block(2)
|
|
nodes[mid].daemon.wait_for_logs(['Ignoring output.*: THEIR_UNILATERAL/THEIR_HTLC'] * 2)
|
|
for _ in range(2):
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
bitcoind.generate_block(1)
|
|
nodes[mid].daemon.wait_for_log('Ignoring output.*: THEIR_UNILATERAL/THEIR_HTLC')
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
|
|
'OUR_UNILATERAL/OUR_HTLC')
|
|
|
|
# At depth 3 we consider them all settled.
|
|
bitcoind.generate_block(3)
|
|
|
|
for n in nodes[mid + 1:]:
|
|
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
|
|
n.rpc.waitsendpay(h, TIMEOUT)
|
|
|
|
# At depth 5, mid+1 can spend HTLC_TIMEOUT_TX output.
|
|
bitcoind.generate_block(1)
|
|
for _ in range(2):
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
|
|
bitcoind.generate_block(1)
|
|
nodes[mid + 1].wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
|
|
|
|
# At depth 100 they're all done.
|
|
bitcoind.generate_block(100)
|
|
nodes[mid].daemon.wait_for_logs(['onchaind complete, forgetting peer'])
|
|
nodes[mid + 1].daemon.wait_for_logs(['onchaind complete, forgetting peer'])
|
|
|
|
# No other channels should have failed.
|
|
for i in range(len(nodes) - 1):
|
|
if i != mid:
|
|
assert only_one(nodes[i].rpc.listpeers(nodes[i + 1].info['id'])['peers'])['connected']
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_permfail_htlc_in(node_factory, bitcoind, executor):
|
|
# Test case where we fail with unsettled incoming HTLC.
|
|
disconnects = ['-WIRE_UPDATE_FULFILL_HTLC', 'permfail']
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l1 = node_factory.get_node(options={'dev-no-reconnect': None}, feerates=(7500, 7500, 7500))
|
|
l2 = node_factory.get_node(disconnect=disconnects)
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 10**6)
|
|
|
|
# This will fail at l2's end.
|
|
t = executor.submit(l1.pay, l2, 200000000)
|
|
|
|
l2.daemon.wait_for_log('dev_disconnect permfail')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('Their unilateral tx, old commit point')
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log('Propose handling OUR_UNILATERAL/THEIR_HTLC by THEIR_HTLC_TIMEOUT_TO_THEM \\(IGNORING\\) after 6 blocks')
|
|
l1.daemon.wait_for_log('Propose handling THEIR_UNILATERAL/OUR_HTLC by OUR_HTLC_TIMEOUT_TO_US (.*) after 6 blocks')
|
|
# l2 then gets preimage, uses it instead of ignoring
|
|
l2.wait_for_onchaind_broadcast('OUR_HTLC_SUCCESS_TX',
|
|
'OUR_UNILATERAL/THEIR_HTLC')
|
|
bitcoind.generate_block(1)
|
|
|
|
# OK, l1 sees l2 fulfill htlc.
|
|
l1.daemon.wait_for_log('THEIR_UNILATERAL/OUR_HTLC gave us preimage')
|
|
l2.daemon.wait_for_log('Propose handling OUR_HTLC_SUCCESS_TX/DELAYED_OUTPUT_TO_US by OUR_DELAYED_RETURN_TO_WALLET .* after 5 blocks')
|
|
bitcoind.generate_block(5)
|
|
|
|
l2.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_HTLC_SUCCESS_TX/DELAYED_OUTPUT_TO_US')
|
|
|
|
t.cancel()
|
|
|
|
# Now, 100 blocks it should be done.
|
|
bitcoind.generate_block(95)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(5)
|
|
l2.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_permfail_htlc_out(node_factory, bitcoind, executor):
|
|
# Test case where we fail with unsettled outgoing HTLC.
|
|
disconnects = ['+WIRE_REVOKE_AND_ACK', 'permfail']
|
|
l1 = node_factory.get_node(options={'dev-no-reconnect': None})
|
|
# Feerates identical so we don't get gratuitous commit to update them
|
|
l2 = node_factory.get_node(disconnect=disconnects, feerates=(7500, 7500, 7500))
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l2.daemon.wait_for_log('openingd-chan#1: Handed peer, entering loop'.format(l1.info['id']))
|
|
l2.fund_channel(l1, 10**6)
|
|
|
|
# This will fail at l2's end.
|
|
t = executor.submit(l2.pay, l1, 200000000)
|
|
|
|
l2.daemon.wait_for_log('dev_disconnect permfail')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('Their unilateral tx, old commit point')
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_logs([
|
|
'Propose handling OUR_UNILATERAL/OUR_HTLC by OUR_HTLC_TIMEOUT_TX \\(.*\\) after 6 blocks',
|
|
'Propose handling OUR_UNILATERAL/DELAYED_OUTPUT_TO_US by OUR_DELAYED_RETURN_TO_WALLET .* after 5 blocks'
|
|
])
|
|
|
|
l1.daemon.wait_for_log('Propose handling THEIR_UNILATERAL/THEIR_HTLC by THEIR_HTLC_TIMEOUT_TO_THEM \\(IGNORING\\) after 6 blocks')
|
|
# l1 then gets preimage, uses it instead of ignoring
|
|
l1.wait_for_onchaind_broadcast('THEIR_HTLC_FULFILL_TO_US',
|
|
'THEIR_UNILATERAL/THEIR_HTLC')
|
|
|
|
# l2 sees l1 fulfill tx.
|
|
bitcoind.generate_block(1)
|
|
|
|
l2.daemon.wait_for_log('OUR_UNILATERAL/OUR_HTLC gave us preimage')
|
|
t.cancel()
|
|
|
|
# l2 can send OUR_DELAYED_RETURN_TO_WALLET after 3 more blocks.
|
|
bitcoind.generate_block(3)
|
|
l2.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# Now, 100 blocks they should be done.
|
|
bitcoind.generate_block(95)
|
|
sync_blockheight(bitcoind, [l1, l2])
|
|
assert not l1.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
l1.daemon.wait_for_log('onchaind complete, forgetting peer')
|
|
sync_blockheight(bitcoind, [l2])
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(3)
|
|
sync_blockheight(bitcoind, [l2])
|
|
assert not l2.daemon.is_in_log('onchaind complete, forgetting peer')
|
|
bitcoind.generate_block(1)
|
|
wait_for(lambda: l2.rpc.listpeers()['peers'] == [])
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_permfail(node_factory, bitcoind):
|
|
l1, l2 = node_factory.line_graph(2)
|
|
|
|
# The funding change should be confirmed and our only output
|
|
assert [o['status'] for o in l1.rpc.listfunds()['outputs']] == ['confirmed']
|
|
l1.pay(l2, 200000000)
|
|
|
|
# Make sure l2 has received sig with 0 htlcs!
|
|
l2.daemon.wait_for_log('Received commit_sig with 1 htlc sigs')
|
|
l2.daemon.wait_for_log('Received commit_sig with 0 htlc sigs')
|
|
|
|
# Make sure l1 has final revocation.
|
|
l1.daemon.wait_for_log('Sending commit_sig with 1 htlc sigs')
|
|
l1.daemon.wait_for_log('Sending commit_sig with 0 htlc sigs')
|
|
l1.daemon.wait_for_log('peer_in WIRE_REVOKE_AND_ACK')
|
|
|
|
# We fail l2, so l1 will reconnect to it.
|
|
l2.rpc.dev_fail(l1.info['id'])
|
|
l2.daemon.wait_for_log('Failing due to dev-fail command')
|
|
l2.wait_for_channel_onchain(l1.info['id'])
|
|
|
|
assert l1.bitcoin.rpc.getmempoolinfo()['size'] == 1
|
|
|
|
# Now grab the close transaction
|
|
closetxid = only_one(l1.bitcoin.rpc.getrawmempool(False))
|
|
|
|
# l2 will send out tx (l1 considers it a transient error)
|
|
bitcoind.generate_block(1)
|
|
|
|
l1.daemon.wait_for_log('Their unilateral tx, old commit point')
|
|
l1.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log(' to ONCHAIN')
|
|
l2.daemon.wait_for_log('Propose handling OUR_UNILATERAL/DELAYED_OUTPUT_TO_US by OUR_DELAYED_RETURN_TO_WALLET (.*) after 5 blocks')
|
|
|
|
wait_for(lambda: only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
|
|
== ['ONCHAIN:Tracking their unilateral close',
|
|
'ONCHAIN:All outputs resolved: waiting 99 more blocks before forgetting channel'])
|
|
|
|
def check_billboard():
|
|
billboard = only_one(l2.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status']
|
|
return (
|
|
len(billboard) == 2
|
|
and billboard[0] == 'ONCHAIN:Tracking our own unilateral close'
|
|
and re.fullmatch(r'ONCHAIN:.* outputs unresolved: in 4 blocks will spend DELAYED_OUTPUT_TO_US \(.*:0\) using OUR_DELAYED_RETURN_TO_WALLET', billboard[1])
|
|
)
|
|
wait_for(check_billboard)
|
|
|
|
# Now, mine 4 blocks so it sends out the spending tx.
|
|
bitcoind.generate_block(4)
|
|
|
|
# onchaind notes to-local payment immediately.
|
|
assert (closetxid, "confirmed") in set([(o['txid'], o['status']) for o in l1.rpc.listfunds()['outputs']])
|
|
|
|
# Restart, should still be confirmed (fails: unwinding blocks erases
|
|
# the confirmation, and we don't re-make it).
|
|
l1.restart()
|
|
wait_for(lambda: (closetxid, "confirmed") in set([(o['txid'], o['status']) for o in l1.rpc.listfunds()['outputs']]))
|
|
|
|
# It should send the to-wallet tx.
|
|
l2.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
|
|
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
|
|
|
|
# 100 after l1 sees tx, it should be done.
|
|
bitcoind.generate_block(95)
|
|
wait_for(lambda: l1.rpc.listpeers()['peers'] == [])
|
|
|
|
wait_for(lambda: only_one(l2.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status'] == [
|
|
'ONCHAIN:Tracking our own unilateral close',
|
|
'ONCHAIN:All outputs resolved: waiting 5 more blocks before forgetting channel'
|
|
])
|
|
|
|
# Now, 100 blocks l2 should be done.
|
|
bitcoind.generate_block(5)
|
|
wait_for(lambda: l2.rpc.listpeers()['peers'] == [])
|
|
|
|
# Only l1 has a direct output since all of l2's outputs are respent (it
|
|
# failed). Also the output should now be listed as confirmed since we
|
|
# generated some more blocks.
|
|
assert (closetxid, "confirmed") in set([(o['txid'], o['status']) for o in l1.rpc.listfunds()['outputs']])
|
|
|
|
# Check that the all the addresses match what we generated ourselves:
|
|
for o in l1.rpc.listfunds()['outputs']:
|
|
txout = bitcoind.rpc.gettxout(o['txid'], o['output'])
|
|
addr = txout['scriptPubKey']['addresses'][0]
|
|
assert(addr == o['address'])
|
|
|
|
addr = l1.bitcoin.getnewaddress()
|
|
l1.rpc.withdraw(addr, "all")
|
|
|
|
|
|
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
|
|
def test_shutdown(node_factory):
|
|
# Fail, in that it will exit before cleanup.
|
|
l1 = node_factory.get_node(may_fail=True)
|
|
if not VALGRIND:
|
|
leaks = l1.rpc.dev_memleak()['leaks']
|
|
if len(leaks):
|
|
raise Exception("Node {} has memory leaks: {}"
|
|
.format(l1.daemon.lightning_dir, leaks))
|
|
l1.rpc.stop()
|
|
|
|
|
|
@flaky
|
|
@unittest.skipIf(not DEVELOPER, "needs to set upfront_shutdown_script")
|
|
def test_option_upfront_shutdown_script(node_factory, bitcoind, executor):
|
|
l1 = node_factory.get_node(start=False)
|
|
# Insist on upfront script we're not going to match.
|
|
l1.daemon.env["DEV_OPENINGD_UPFRONT_SHUTDOWN_SCRIPT"] = "76a91404b61f7dc1ea0dc99424464cc4064dc564d91e8988ac"
|
|
l1.start()
|
|
|
|
l2 = node_factory.get_node()
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 1000000, False)
|
|
|
|
# This will block, as l12 will send an error but l2 will retry.
|
|
fut = executor.submit(l1.rpc.close, l2.info['id'])
|
|
|
|
# l2 will close unilaterally when it dislikes shutdown script.
|
|
l1.daemon.wait_for_log(r'scriptpubkey .* is not as agreed upfront \(76a91404b61f7dc1ea0dc99424464cc4064dc564d91e8988ac\)')
|
|
|
|
# Clear channel.
|
|
wait_for(lambda: len(bitcoind.rpc.getrawmempool()) != 0)
|
|
bitcoind.generate_block(1)
|
|
fut.result(TIMEOUT)
|
|
wait_for(lambda: [c['state'] for c in only_one(l1.rpc.listpeers()['peers'])['channels']] == ['ONCHAIN'])
|
|
wait_for(lambda: [c['state'] for c in only_one(l2.rpc.listpeers()['peers'])['channels']] == ['ONCHAIN'])
|
|
|
|
# Works when l2 closes channel, too.
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.fund_channel(l2, 1000000, False)
|
|
|
|
l2.rpc.close(l1.info['id'])
|
|
|
|
# l2 will close unilaterally when it dislikes shutdown script.
|
|
l1.daemon.wait_for_log(r'scriptpubkey .* is not as agreed upfront \(76a91404b61f7dc1ea0dc99424464cc4064dc564d91e8988ac\)')
|
|
|
|
# Clear channel.
|
|
wait_for(lambda: len(bitcoind.rpc.getrawmempool()) != 0)
|
|
bitcoind.generate_block(1)
|
|
wait_for(lambda: [c['state'] for c in only_one(l1.rpc.listpeers()['peers'])['channels']] == ['ONCHAIN', 'ONCHAIN'])
|
|
wait_for(lambda: [c['state'] for c in only_one(l2.rpc.listpeers()['peers'])['channels']] == ['ONCHAIN', 'ONCHAIN'])
|
|
|
|
# Figure out what address it will try to use.
|
|
keyidx = int(l1.db_query("SELECT intval FROM vars WHERE name='bip32_max_index';")[0]['intval'])
|
|
|
|
# Expect 1 for change address, 1 for the channel final address,
|
|
# which are discarded as the 'scratch' tx that the fundchannel
|
|
# plugin makes, plus 1 for the funding address of the actual
|
|
# funding tx.
|
|
addr = l1.rpc.call('dev-listaddrs', [keyidx + 3])['addresses'][-1]
|
|
|
|
# Now, if we specify upfront and it's OK, all good.
|
|
l1.stop()
|
|
# We need to prepend the segwit version (0) and push opcode (14).
|
|
l1.daemon.env["DEV_OPENINGD_UPFRONT_SHUTDOWN_SCRIPT"] = '0014' + addr['bech32_redeemscript']
|
|
l1.start()
|
|
|
|
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
|
|
l1.rpc.fundchannel(l2.info['id'], 1000000)
|
|
l1.rpc.close(l2.info['id'])
|
|
wait_for(lambda: sorted([c['state'] for c in only_one(l1.rpc.listpeers()['peers'])['channels']]) == ['CLOSINGD_COMPLETE', 'ONCHAIN', 'ONCHAIN'])
|
|
|