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from fixtures import * # noqa: F401,F403
from lightning import RpcError
from utils import only_one, sync_blockheight, wait_for, DEVELOPER, TIMEOUT, VALGRIND, SLOW_MACHINE
import queue
import pytest
import re
import threading
import unittest
@unittest.skipIf(not DEVELOPER, "Too slow without --dev-bitcoind-poll")
def test_closing(node_factory, bitcoind):
l1, l2 = node_factory.line_graph(2)
chan = l1.get_channel_scid(l2)
l1.pay(l2, 200000000)
assert bitcoind.rpc.getmempoolinfo()['size'] == 0
billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
assert billboard == ['CHANNELD_NORMAL:Funding transaction locked.']
billboard = only_one(l2.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status']
assert billboard == ['CHANNELD_NORMAL:Funding transaction locked.']
bitcoind.generate_block(5)
# Only wait for the channels to activate with DEVELOPER=1,
# otherwise it's going to take too long because of the missing
# --dev-broadcast-interval
if DEVELOPER:
wait_for(lambda: len(l1.getactivechannels()) == 2)
wait_for(lambda: len(l2.getactivechannels()) == 2)
billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
# This may either be from a local_update or an announce, so just
# check for the substring
assert 'CHANNELD_NORMAL:Funding transaction locked.' in billboard[0]
# This should return with an error, then close.
with pytest.raises(RpcError, match=r'Channel close negotiation not finished'):
l1.rpc.close(chan, False, 0)
l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
l2.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
# And should put closing into mempool.
l1.daemon.wait_for_log('sendrawtx exit 0')
l2.daemon.wait_for_log('sendrawtx exit 0')
# Both nodes should have disabled the channel in their view
wait_for(lambda: len(l1.getactivechannels()) == 0)
wait_for(lambda: len(l2.getactivechannels()) == 0)
assert bitcoind.rpc.getmempoolinfo()['size'] == 1
# Now grab the close transaction
closetxid = only_one(bitcoind.rpc.getrawmempool(False))
billboard = only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status']
assert billboard == ['CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of 5430 satoshi']
bitcoind.generate_block(1)
l1.daemon.wait_for_log(r'Owning output .* txid %s' % closetxid)
l2.daemon.wait_for_log(r'Owning output .* txid %s' % closetxid)
# Make sure both nodes have grabbed their close tx funds
assert closetxid in set([o['txid'] for o in l1.rpc.listfunds()['outputs']])
assert closetxid in set([o['txid'] for o in l2.rpc.listfunds()['outputs']])
wait_for(lambda: only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status'] == [
'CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of 5430 satoshi',
'ONCHAIN:Tracking mutual close transaction',
'ONCHAIN:All outputs resolved: waiting 99 more blocks before forgetting channel'
])
bitcoind.generate_block(9)
wait_for(lambda: only_one(l1.rpc.listpeers(l2.info['id'])['peers'][0]['channels'])['status'] == [
'CLOSINGD_SIGEXCHANGE:We agreed on a closing fee of 5430 satoshi',
'ONCHAIN:Tracking mutual close transaction',
'ONCHAIN:All outputs resolved: waiting 90 more blocks before forgetting channel'
])
# Make sure both have forgotten about it
bitcoind.generate_block(90)
wait_for(lambda: len(l1.rpc.listchannels()['channels']) == 0)
wait_for(lambda: len(l2.rpc.listchannels()['channels']) == 0)
def test_closing_while_disconnected(node_factory, bitcoind):
l1, l2 = node_factory.line_graph(2, opts={'may_reconnect': True})
chan = l1.get_channel_scid(l2)
l1.pay(l2, 200000000)
l2.stop()
# The close should still be triggered afterwards.
with pytest.raises(RpcError, match=r'Channel close negotiation not finished'):
l1.rpc.close(chan, False, 0)
l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
l2.start()
l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
# And should put closing into mempool.
l1.daemon.wait_for_log('sendrawtx exit 0')
l2.daemon.wait_for_log('sendrawtx exit 0')
bitcoind.generate_block(101)
wait_for(lambda: len(l1.rpc.listchannels()['channels']) == 0)
wait_for(lambda: len(l2.rpc.listchannels()['channels']) == 0)
def test_closing_id(node_factory):
"""Test closing using peer ID and full channel ID
"""
l1, l2 = node_factory.get_nodes(2)
# Close by full channel ID.
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
l1.fund_channel(l2, 10**6)
cid = l2.rpc.listpeers()['peers'][0]['channels'][0]['channel_id']
l2.rpc.close(cid)
wait_for(lambda: not only_one(l1.rpc.listpeers(l2.info['id'])['peers'])['connected'])
wait_for(lambda: not only_one(l2.rpc.listpeers(l1.info['id'])['peers'])['connected'])
# Close by peer ID.
l2.rpc.connect(l1.info['id'], 'localhost', l1.port)
l1.daemon.wait_for_log("Handed peer, entering loop")
l2.fund_channel(l1, 10**6)
pid = l1.info['id']
l2.rpc.close(pid)
wait_for(lambda: not only_one(l1.rpc.listpeers(l2.info['id'])['peers'])['connected'])
wait_for(lambda: not only_one(l2.rpc.listpeers(l1.info['id'])['peers'])['connected'])
@unittest.skipIf(not DEVELOPER, "needs dev-rescan-outputs")
def test_closing_torture(node_factory, executor, bitcoind):
l1, l2 = node_factory.get_nodes(2)
amount = 10**6
# Before the fix was applied, 15 would often pass.
# However, increasing the number of tries would
# take longer in VALGRIND mode, triggering a CI
# failure since the test does not print any
# output.
# On my laptop, VALGRIND is about 4x slower than native, hence
# the approximations below:
iterations = 50
if VALGRIND:
iterations //= 4
if SLOW_MACHINE:
iterations //= 2
for i in range(iterations):
# Reduce probability that spurious sendrawtx error will occur
l1.rpc.dev_rescan_outputs()
# Create a channel.
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
l1.fund_channel(l2, amount)
scid = l1.get_channel_scid(l2)
# Get it confirmed.
l1.bitcoin.generate_block(6)
# Wait for it to go to CHANNELD_NORMAL
l1.wait_channel_active(scid)
l2.wait_channel_active(scid)
# Start closers: can take a long time under valgrind!
c1 = executor.submit(l1.rpc.close, l2.info['id'], False, 60)
c2 = executor.submit(l2.rpc.close, l1.info['id'], False, 60)
# Wait for close to finish
c1.result(TIMEOUT)
c2.result(TIMEOUT)
wait_for(lambda: len(bitcoind.rpc.getrawmempool(False)) == 1)
# Get close confirmed
l1.bitcoin.generate_block(100)
wait_for(lambda: len(l1.rpc.listpeers()['peers']) == 0)
wait_for(lambda: len(l2.rpc.listpeers()['peers']) == 0)
@unittest.skipIf(SLOW_MACHINE and VALGRIND, "slow test")
def test_closing_different_fees(node_factory, bitcoind, executor):
l1 = node_factory.get_node()
# Default feerate = 15000/7500/1000
# It will start at the second number, accepting anything above the first.
feerates = [[20000, 15000, 7400], [8000, 1001, 100]]
amounts = [0, 545999, 546000]
num_peers = len(feerates) * len(amounts)
addr = l1.rpc.newaddr()['address']
bitcoind.rpc.sendtoaddress(addr, 1)
numfunds = len(l1.rpc.listfunds()['outputs'])
bitcoind.generate_block(1)
wait_for(lambda: len(l1.rpc.listfunds()['outputs']) > numfunds)
# Create them in a batch, for speed!
peers = []
for feerate in feerates:
for amount in amounts:
p = node_factory.get_node(feerates=feerate)
p.feerate = feerate
p.amount = amount
l1.rpc.connect(p.info['id'], 'localhost', p.port)
peers.append(p)
for p in peers:
p.channel = l1.rpc.fundchannel(p.info['id'], 10**6)['channel_id']
# Technically, this is async to fundchannel returning.
l1.daemon.wait_for_log('sendrawtx exit 0')
bitcoind.generate_block(6)
# Now wait for them all to hit normal state, do payments
l1.daemon.wait_for_logs(['update for channel .* now ACTIVE'] * num_peers
+ ['to CHANNELD_NORMAL'] * num_peers)
for p in peers:
if p.amount != 0:
l1.pay(p, 100000000)
# Now close all channels
# All closes occur in parallel, and on Travis,
# ALL those lightningd are running on a single core,
# so increase the timeout so that this test will pass
# when valgrind is enabled.
# (close timeout defaults to 30 as of this writing)
closes = [executor.submit(l1.rpc.close, p.channel, False, 90) for p in peers]
for c in closes:
c.result(90)
# close does *not* wait for the sendrawtransaction, so do that!
# Note that since they disagree on the ideal fee, they may conflict
# (first one in will win), so we cannot look at logs, we need to
# wait for mempool.
wait_for(lambda: bitcoind.rpc.getmempoolinfo()['size'] == num_peers)
bitcoind.generate_block(1)
for p in peers:
p.daemon.wait_for_log(' to ONCHAIN')
wait_for(lambda: 'ONCHAIN:Tracking mutual close transaction' in only_one(p.rpc.listpeers(l1.info['id'])['peers'][0]['channels'])['status'])
l1.daemon.wait_for_logs([' to ONCHAIN'] * num_peers)
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
def test_closing_negotiation_reconnect(node_factory, bitcoind):
disconnects = ['-WIRE_CLOSING_SIGNED',
'@WIRE_CLOSING_SIGNED',
'+WIRE_CLOSING_SIGNED']
l1 = node_factory.get_node(disconnect=disconnects, may_reconnect=True)
l2 = node_factory.get_node(may_reconnect=True)
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
chan = l1.fund_channel(l2, 10**6)
l1.pay(l2, 200000000)
assert bitcoind.rpc.getmempoolinfo()['size'] == 0
# This should return with an error, then close.
with pytest.raises(RpcError, match=r'Channel close negotiation not finished'):
l1.rpc.close(chan, False, 0)
l1.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
l2.daemon.wait_for_log(' to CHANNELD_SHUTTING_DOWN')
l1.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
l2.daemon.wait_for_log(' to CLOSINGD_SIGEXCHANGE')
# And should put closing into mempool (happens async, so
# CLOSINGD_COMPLETE may come first).
l1.daemon.wait_for_logs(['sendrawtx exit 0', ' to CLOSINGD_COMPLETE'])
l2.daemon.wait_for_logs(['sendrawtx exit 0', ' to CLOSINGD_COMPLETE'])
assert bitcoind.rpc.getmempoolinfo()['size'] == 1
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
def test_penalty_inhtlc(node_factory, bitcoind, executor):
"""Test penalty transaction with an incoming HTLC"""
# We suppress each one after first commit; HTLC gets added not fulfilled.
# Feerates identical so we don't get gratuitous commit to update them
l1 = node_factory.get_node(disconnect=['=WIRE_COMMITMENT_SIGNED-nocommit'], may_fail=True, feerates=(7500, 7500, 7500))
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('UPDATE WIRE_CHANNEL_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.
assert sum(o['value'] for o in outputs) < 10**6
assert sum(o['value'] for o in outputs) > 10**6 - 15000
@unittest.skipIf(not DEVELOPER, "needs DEVELOPER=1")
def test_penalty_outhtlc(node_factory, bitcoind, executor):
"""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))
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.
assert sum(o['value'] for o in outputs) < 10**6
assert sum(o['value'] for o in outputs) > 10**6 - 15000
@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()['address'], '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()['address'], '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.isAlive()
# 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.
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.
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']])
addr = l1.bitcoin.rpc.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()