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#!/usr/bin/env python3
# Copyright (c) 2015 The Bitcoin Core developers
# Distributed under the MIT software license, see the accompanying
# file COPYING or http://www.opensource.org/licenses/mit-license.php.
#
# Test replace-by-fee
#
import os
import sys
# Add python-bitcoinlib to module search path:
sys.path.append(os.path.join(os.path.dirname(os.path.abspath(__file__)), "python-bitcoinlib"))
import unittest
import bitcoin
bitcoin.SelectParams('regtest')
import bitcoin.rpc
from bitcoin.core import *
from bitcoin.core.script import *
from bitcoin.wallet import *
MAX_REPLACEMENT_LIMIT = 100
class Test_ReplaceByFee(unittest.TestCase):
proxy = None
@classmethod
def setUpClass(cls):
if cls.proxy is None:
cls.proxy = bitcoin.rpc.Proxy()
@classmethod
def tearDownClass(cls):
# Make sure mining works
mempool_size = 1
while mempool_size:
cls.proxy.call('generate',1)
new_mempool_size = len(cls.proxy.getrawmempool())
# It's possible to get stuck in a loop here if the mempool has
# transactions that can't be mined.
assert(new_mempool_size != mempool_size)
mempool_size = new_mempool_size
def make_txout(self, amount, scriptPubKey=CScript([1])):
"""Create a txout with a given amount and scriptPubKey
Mines coins as needed.
"""
fee = 1*COIN
while self.proxy.getbalance() < amount + fee:
self.proxy.call('generate', 100)
addr = P2SHBitcoinAddress.from_redeemScript(CScript([]))
txid = self.proxy.sendtoaddress(addr, amount + fee)
tx1 = self.proxy.getrawtransaction(txid)
i = None
for i, txout in enumerate(tx1.vout):
if txout.scriptPubKey == addr.to_scriptPubKey():
break
assert i is not None
tx2 = CTransaction([CTxIn(COutPoint(txid, i), CScript([1, CScript([])]), nSequence=0)],
[CTxOut(amount, scriptPubKey)])
tx2_txid = self.proxy.sendrawtransaction(tx2, True)
return COutPoint(tx2_txid, 0)
def test_simple_doublespend(self):
"""Simple doublespend"""
tx0_outpoint = self.make_txout(1.1*COIN)
tx1a = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(1*COIN, CScript([b'a']))])
tx1a_txid = self.proxy.sendrawtransaction(tx1a, True)
# Should fail because we haven't changed the fee
tx1b = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(1*COIN, CScript([b'b']))])
try:
tx1b_txid = self.proxy.sendrawtransaction(tx1b, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26) # insufficient fee
else:
self.fail()
# Extra 0.1 BTC fee
tx1b = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(0.9*COIN, CScript([b'b']))])
tx1b_txid = self.proxy.sendrawtransaction(tx1b, True)
# tx1a is in fact replaced
with self.assertRaises(IndexError):
self.proxy.getrawtransaction(tx1a_txid)
self.assertEqual(tx1b, self.proxy.getrawtransaction(tx1b_txid))
def test_doublespend_chain(self):
"""Doublespend of a long chain"""
initial_nValue = 50*COIN
tx0_outpoint = self.make_txout(initial_nValue)
prevout = tx0_outpoint
remaining_value = initial_nValue
chain_txids = []
while remaining_value > 10*COIN:
remaining_value -= 1*COIN
tx = CTransaction([CTxIn(prevout, nSequence=0)],
[CTxOut(remaining_value, CScript([1]))])
txid = self.proxy.sendrawtransaction(tx, True)
chain_txids.append(txid)
prevout = COutPoint(txid, 0)
# Whether the double-spend is allowed is evaluated by including all
# child fees - 40 BTC - so this attempt is rejected.
dbl_tx = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(initial_nValue - 30*COIN, CScript([1]))])
try:
self.proxy.sendrawtransaction(dbl_tx, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26) # insufficient fee
else:
self.fail()
# Accepted with sufficient fee
dbl_tx = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(1*COIN, CScript([1]))])
self.proxy.sendrawtransaction(dbl_tx, True)
for doublespent_txid in chain_txids:
with self.assertRaises(IndexError):
self.proxy.getrawtransaction(doublespent_txid)
def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions"""
initial_nValue = 50*COIN
tx0_outpoint = self.make_txout(initial_nValue)
def branch(prevout, initial_value, max_txs, *, tree_width=5, fee=0.0001*COIN, _total_txs=None):
if _total_txs is None:
_total_txs = [0]
if _total_txs[0] >= max_txs:
return
txout_value = (initial_value - fee) // tree_width
if txout_value < fee:
return
vout = [CTxOut(txout_value, CScript([i+1]))
for i in range(tree_width)]
tx = CTransaction([CTxIn(prevout, nSequence=0)],
vout)
self.assertTrue(len(tx.serialize()) < 100000)
txid = self.proxy.sendrawtransaction(tx, True)
yield tx
_total_txs[0] += 1
for i, txout in enumerate(tx.vout):
yield from branch(COutPoint(txid, i), txout_value,
max_txs,
tree_width=tree_width, fee=fee,
_total_txs=_total_txs)
fee = 0.0001*COIN
n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
self.assertEqual(len(tree_txs), n)
# Attempt double-spend, will fail because too little fee paid
dbl_tx = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(initial_nValue - fee*n, CScript([1]))])
try:
self.proxy.sendrawtransaction(dbl_tx, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26) # insufficient fee
else:
self.fail()
# 1 BTC fee is enough
dbl_tx = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(initial_nValue - fee*n - 1*COIN, CScript([1]))])
self.proxy.sendrawtransaction(dbl_tx, True)
for tx in tree_txs:
with self.assertRaises(IndexError):
self.proxy.getrawtransaction(tx.GetHash())
# Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT, MAX_REPLACEMENT_LIMIT*2):
fee = 0.0001*COIN
tx0_outpoint = self.make_txout(initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
self.assertEqual(len(tree_txs), n)
dbl_tx = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(initial_nValue - fee*n, CScript([1]))])
try:
self.proxy.sendrawtransaction(dbl_tx, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26)
else:
self.fail()
for tx in tree_txs:
self.proxy.getrawtransaction(tx.GetHash())
def test_replacement_feeperkb(self):
"""Replacement requires overall fee-per-KB to be higher"""
tx0_outpoint = self.make_txout(1.1*COIN)
tx1a = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(1*COIN, CScript([b'a']))])
tx1a_txid = self.proxy.sendrawtransaction(tx1a, True)
# Higher fee, but the fee per KB is much lower, so the replacement is
# rejected.
tx1b = CTransaction([CTxIn(tx0_outpoint, nSequence=0)],
[CTxOut(0.001*COIN,
CScript([b'a'*999000]))])
try:
tx1b_txid = self.proxy.sendrawtransaction(tx1b, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26) # insufficient fee
else:
self.fail()
def test_spends_of_conflicting_outputs(self):
"""Replacements that spend conflicting tx outputs are rejected"""
utxo1 = self.make_txout(1.2*COIN)
utxo2 = self.make_txout(3.0*COIN)
tx1a = CTransaction([CTxIn(utxo1, nSequence=0)],
[CTxOut(1.1*COIN, CScript([b'a']))])
tx1a_txid = self.proxy.sendrawtransaction(tx1a, True)
# Direct spend an output of the transaction we're replacing.
tx2 = CTransaction([CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0),
CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)],
tx1a.vout)
try:
tx2_txid = self.proxy.sendrawtransaction(tx2, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26)
else:
self.fail()
# Spend tx1a's output to test the indirect case.
tx1b = CTransaction([CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)],
[CTxOut(1.0*COIN, CScript([b'a']))])
tx1b_txid = self.proxy.sendrawtransaction(tx1b, True)
tx2 = CTransaction([CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0),
CTxIn(COutPoint(tx1b_txid, 0))],
tx1a.vout)
try:
tx2_txid = self.proxy.sendrawtransaction(tx2, True)
except bitcoin.rpc.JSONRPCException as exp:
self.assertEqual(exp.error['code'], -26)
else:
self.fail()
if __name__ == '__main__':
unittest.main()