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Fix negative amounts assertion

Change amounts could be negative after deducting the cost of the extra
change output; floor them at zero.  Move the assertion to the main
code.  Simplify rounding logic.
283
Neil Booth 9 years ago
parent
commit
5c3a6db445
  1. 16
      lib/coinchooser.py

16
lib/coinchooser.py

@ -61,11 +61,12 @@ class CoinChooserBase(PrintError):
def change_amounts(self, tx, count, fee_estimator, dust_threshold):
# The amount left after adding 1 change output
return [tx.get_fee() - fee_estimator(1)]
return [max(0, tx.get_fee() - fee_estimator(1))]
def change_outputs(self, tx, change_addrs, fee_estimator, dust_threshold):
amounts = self.change_amounts(tx, len(change_addrs), fee_estimator,
dust_threshold)
assert min(amounts) >= 0
# If change is above dust threshold after accounting for the
# size of the change output, add it to the transaction.
dust = sum(amount for amount in amounts if amount < dust_threshold)
@ -228,8 +229,8 @@ class CoinChooserPrivacy(CoinChooserRandom):
# Use N change outputs
for n in range(1, count + 1):
# How much is left if we add this many change outputs?
change_amount = tx.get_fee() - fee_estimator(n)
if change_amount // n < max_change:
change_amount = max(0, tx.get_fee() - fee_estimator(n))
if change_amount // n <= max_change:
break
# Get a handle on the precision of the output amounts; round our
@ -257,16 +258,11 @@ class CoinChooserPrivacy(CoinChooserRandom):
# Last change output. Round down to maximum precision but lose
# no more than 100 satoshis to fees (2dp)
amount = remaining
N = min(2, zeroes[0])
if N:
amount = int(round(amount, -N))
if amount > remaining:
amount -= pow(10, N)
N = pow(10, min(2, zeroes[0]))
amount = (remaining // N) * N
amounts.append(amount)
assert sum(amounts) <= change_amount
assert min(amounts) >= 0
return amounts

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