tFix negative amounts assertion - electrum - Electrum Bitcoin wallet
HTML git clone https://git.parazyd.org/electrum
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DIR Files
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---
DIR commit 5c3a6db445ef65d960b8140a57fa3a75a6e11aee
DIR parent ee4ccd9b1be300f4e4a0801effeec2cb4a75df04
HTML Author: Neil Booth <kyuupichan@gmail.com>
Date: Sat, 19 Dec 2015 13:44:12 +0900
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.
Diffstat:
M lib/coinchooser.py | 16 ++++++----------
1 file changed, 6 insertions(+), 10 deletions(-)
---
DIR diff --git a/lib/coinchooser.py b/lib/coinchooser.py
t@@ -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)
t@@ -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
t@@ -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