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kevacoin/test/functional/replace-by-fee.py

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#!/usr/bin/env python3
# Copyright (c) 2014-2016 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 the RBF code."""
from test_framework.test_framework import BitcoinTestFramework
from test_framework.util import *
from test_framework.script import *
from test_framework.mininode import *
MAX_REPLACEMENT_LIMIT = 100
def txToHex(tx):
return bytes_to_hex_str(tx.serialize())
def make_utxo(node, amount, confirmed=True, scriptPubKey=CScript([1])):
"""Create a txout with a given amount and scriptPubKey
Mines coins as needed.
confirmed - txouts created will be confirmed in the blockchain;
unconfirmed otherwise.
"""
fee = 1*COIN
while node.getbalance() < satoshi_round((amount + fee)/COIN):
node.generate(100)
new_addr = node.getnewaddress()
txid = node.sendtoaddress(new_addr, satoshi_round((amount+fee)/COIN))
tx1 = node.getrawtransaction(txid, 1)
txid = int(txid, 16)
i = None
for i, txout in enumerate(tx1['vout']):
if txout['scriptPubKey']['addresses'] == [new_addr]:
break
assert i is not None
tx2 = CTransaction()
tx2.vin = [CTxIn(COutPoint(txid, i))]
tx2.vout = [CTxOut(amount, scriptPubKey)]
tx2.rehash()
signed_tx = node.signrawtransaction(txToHex(tx2))
txid = node.sendrawtransaction(signed_tx['hex'], True)
# If requested, ensure txouts are confirmed.
if confirmed:
mempool_size = len(node.getrawmempool())
while mempool_size > 0:
node.generate(1)
new_size = len(node.getrawmempool())
# Error out if we have something stuck in the mempool, as this
# would likely be a bug.
assert(new_size < mempool_size)
mempool_size = new_size
return COutPoint(int(txid, 16), 0)
class ReplaceByFeeTest(BitcoinTestFramework):
def set_test_params(self):
self.num_nodes = 2
self.extra_args= [["-maxorphantx=1000",
"-whitelist=127.0.0.1",
"-limitancestorcount=50",
"-limitancestorsize=101",
"-limitdescendantcount=200",
"-limitdescendantsize=101"],
["-mempoolreplacement=0"]]
def run_test(self):
# Leave IBD
self.nodes[0].generate(1)
make_utxo(self.nodes[0], 1*COIN)
# Ensure nodes are synced
self.sync_all()
self.log.info("Running test simple doublespend...")
self.test_simple_doublespend()
self.log.info("Running test doublespend chain...")
self.test_doublespend_chain()
self.log.info("Running test doublespend tree...")
self.test_doublespend_tree()
self.log.info("Running test replacement feeperkb...")
self.test_replacement_feeperkb()
self.log.info("Running test spends of conflicting outputs...")
self.test_spends_of_conflicting_outputs()
self.log.info("Running test new unconfirmed inputs...")
self.test_new_unconfirmed_inputs()
self.log.info("Running test too many replacements...")
self.test_too_many_replacements()
self.log.info("Running test opt-in...")
self.test_opt_in()
self.log.info("Running test RPC...")
self.test_rpc()
self.log.info("Running test prioritised transactions...")
self.test_prioritised_transactions()
self.log.info("Passed")
def test_simple_doublespend(self):
"""Simple doublespend"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# make_utxo may have generated a bunch of blocks, so we need to sync
# before we can spend the coins generated, or else the resulting
# transactions might not be accepted by our peers.
self.sync_all()
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
self.sync_all()
# Should fail because we haven't changed the fee
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(1*COIN, CScript([b'b']))]
tx1b_hex = txToHex(tx1b)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
# This will raise an exception due to transaction replacement being disabled
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[1].sendrawtransaction, tx1b_hex, True)
# Extra 0.1 BTC fee
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.9*COIN), CScript([b'b']))]
tx1b_hex = txToHex(tx1b)
# Replacement still disabled even with "enough fee"
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[1].sendrawtransaction, tx1b_hex, True)
# Works when enabled
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
mempool = self.nodes[0].getrawmempool()
assert (tx1a_txid not in mempool)
assert (tx1b_txid in mempool)
assert_equal(tx1b_hex, self.nodes[0].getrawtransaction(tx1b_txid))
# Second node is running mempoolreplacement=0, will not replace originally-seen txn
mempool = self.nodes[1].getrawmempool()
assert tx1a_txid in mempool
assert tx1b_txid not in mempool
def test_doublespend_chain(self):
"""Doublespend of a long chain"""
initial_nValue = 50*COIN
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
prevout = tx0_outpoint
remaining_value = initial_nValue
chain_txids = []
while remaining_value > 10*COIN:
remaining_value -= 1*COIN
tx = CTransaction()
tx.vin = [CTxIn(prevout, nSequence=0)]
tx.vout = [CTxOut(remaining_value, CScript([1]))]
tx_hex = txToHex(tx)
txid = self.nodes[0].sendrawtransaction(tx_hex, True)
chain_txids.append(txid)
prevout = COutPoint(int(txid, 16), 0)
# Whether the double-spend is allowed is evaluated by including all
# child fees - 40 BTC - so this attempt is rejected.
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - 30*COIN, CScript([1]))]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
# Accepted with sufficient fee
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(1*COIN, CScript([1]))]
dbl_tx_hex = txToHex(dbl_tx)
self.nodes[0].sendrawtransaction(dbl_tx_hex, True)
mempool = self.nodes[0].getrawmempool()
for doublespent_txid in chain_txids:
assert(doublespent_txid not in mempool)
def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions"""
initial_nValue = 50*COIN
tx0_outpoint = make_utxo(self.nodes[0], 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()
tx.vin = [CTxIn(prevout, nSequence=0)]
tx.vout = vout
tx_hex = txToHex(tx)
assert(len(tx.serialize()) < 100000)
txid = self.nodes[0].sendrawtransaction(tx_hex, True)
yield tx
_total_txs[0] += 1
txid = int(txid, 16)
for i, txout in enumerate(tx.vout):
for x in branch(COutPoint(txid, i), txout_value,
max_txs,
tree_width=tree_width, fee=fee,
_total_txs=_total_txs):
yield x
fee = int(0.0001*COIN)
n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
# Attempt double-spend, will fail because too little fee paid
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - fee*n, CScript([1]))]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
# 1 BTC fee is enough
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - fee*n - 1*COIN, CScript([1]))]
dbl_tx_hex = txToHex(dbl_tx)
self.nodes[0].sendrawtransaction(dbl_tx_hex, True)
mempool = self.nodes[0].getrawmempool()
for tx in tree_txs:
tx.rehash()
assert (tx.hash not in mempool)
# Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT+1, MAX_REPLACEMENT_LIMIT*2):
fee = int(0.0001*COIN)
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - 2*fee*n, CScript([1]))]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
for tx in tree_txs:
tx.rehash()
self.nodes[0].getrawtransaction(tx.hash)
def test_replacement_feeperkb(self):
"""Replacement requires fee-per-KB to be higher"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1a_hex = txToHex(tx1a)
self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Higher fee, but the fee per KB is much lower, so the replacement is
# rejected.
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.001*COIN), CScript([b'a'*999000]))]
tx1b_hex = txToHex(tx1b)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
def test_spends_of_conflicting_outputs(self):
"""Replacements that spend conflicting tx outputs are rejected"""
utxo1 = make_utxo(self.nodes[0], int(1.2*COIN))
utxo2 = make_utxo(self.nodes[0], 3*COIN)
tx1a = CTransaction()
tx1a.vin = [CTxIn(utxo1, nSequence=0)]
tx1a.vout = [CTxOut(int(1.1*COIN), CScript([b'a']))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
tx1a_txid = int(tx1a_txid, 16)
# Direct spend an output of the transaction we're replacing.
tx2 = CTransaction()
tx2.vin = [CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0)]
tx2.vin.append(CTxIn(COutPoint(tx1a_txid, 0), nSequence=0))
tx2.vout = tx1a.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "bad-txns-spends-conflicting-tx", self.nodes[0].sendrawtransaction, tx2_hex, True)
# Spend tx1a's output to test the indirect case.
tx1b = CTransaction()
tx1b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
tx1b.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1b_hex = txToHex(tx1b)
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
tx1b_txid = int(tx1b_txid, 16)
tx2 = CTransaction()
tx2.vin = [CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0),
CTxIn(COutPoint(tx1b_txid, 0))]
tx2.vout = tx1a.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "bad-txns-spends-conflicting-tx", self.nodes[0].sendrawtransaction, tx2_hex, True)
def test_new_unconfirmed_inputs(self):
"""Replacements that add new unconfirmed inputs are rejected"""
confirmed_utxo = make_utxo(self.nodes[0], int(1.1*COIN))
unconfirmed_utxo = make_utxo(self.nodes[0], int(0.1*COIN), False)
tx1 = CTransaction()
tx1.vin = [CTxIn(confirmed_utxo)]
tx1.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1_hex = txToHex(tx1)
self.nodes[0].sendrawtransaction(tx1_hex, True)
tx2 = CTransaction()
tx2.vin = [CTxIn(confirmed_utxo), CTxIn(unconfirmed_utxo)]
tx2.vout = tx1.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "replacement-adds-unconfirmed", self.nodes[0].sendrawtransaction, tx2_hex, True)
def test_too_many_replacements(self):
"""Replacements that evict too many transactions are rejected"""
# Try directly replacing more than MAX_REPLACEMENT_LIMIT
# transactions
# Start by creating a single transaction with many outputs
initial_nValue = 10*COIN
utxo = make_utxo(self.nodes[0], initial_nValue)
fee = int(0.0001*COIN)
split_value = int((initial_nValue-fee)/(MAX_REPLACEMENT_LIMIT+1))
outputs = []
for i in range(MAX_REPLACEMENT_LIMIT+1):
outputs.append(CTxOut(split_value, CScript([1])))
splitting_tx = CTransaction()
splitting_tx.vin = [CTxIn(utxo, nSequence=0)]
splitting_tx.vout = outputs
splitting_tx_hex = txToHex(splitting_tx)
txid = self.nodes[0].sendrawtransaction(splitting_tx_hex, True)
txid = int(txid, 16)
# Now spend each of those outputs individually
for i in range(MAX_REPLACEMENT_LIMIT+1):
tx_i = CTransaction()
tx_i.vin = [CTxIn(COutPoint(txid, i), nSequence=0)]
tx_i.vout = [CTxOut(split_value-fee, CScript([b'a']))]
tx_i_hex = txToHex(tx_i)
self.nodes[0].sendrawtransaction(tx_i_hex, True)
# Now create doublespend of the whole lot; should fail.
# Need a big enough fee to cover all spending transactions and have
# a higher fee rate
double_spend_value = (split_value-100*fee)*(MAX_REPLACEMENT_LIMIT+1)
inputs = []
for i in range(MAX_REPLACEMENT_LIMIT+1):
inputs.append(CTxIn(COutPoint(txid, i), nSequence=0))
double_tx = CTransaction()
double_tx.vin = inputs
double_tx.vout = [CTxOut(double_spend_value, CScript([b'a']))]
double_tx_hex = txToHex(double_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements", self.nodes[0].sendrawtransaction, double_tx_hex, True)
# If we remove an input, it should pass
double_tx = CTransaction()
double_tx.vin = inputs[0:-1]
double_tx.vout = [CTxOut(double_spend_value, CScript([b'a']))]
double_tx_hex = txToHex(double_tx)
self.nodes[0].sendrawtransaction(double_tx_hex, True)
def test_opt_in(self):
"""Replacing should only work if orig tx opted in"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# Create a non-opting in transaction
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0xffffffff)]
tx1a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Shouldn't be able to double-spend
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.9*COIN), CScript([b'b']))]
tx1b_hex = txToHex(tx1b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[0].sendrawtransaction, tx1b_hex, True)
tx1_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# Create a different non-opting in transaction
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0xfffffffe)]
tx2a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx2a_hex = txToHex(tx2a)
tx2a_txid = self.nodes[0].sendrawtransaction(tx2a_hex, True)
# Still shouldn't be able to double-spend
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(0.9*COIN), CScript([b'b']))]
tx2b_hex = txToHex(tx2b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[0].sendrawtransaction, tx2b_hex, True)
# Now create a new transaction that spends from tx1a and tx2a
# opt-in on one of the inputs
# Transaction should be replaceable on either input
tx1a_txid = int(tx1a_txid, 16)
tx2a_txid = int(tx2a_txid, 16)
tx3a = CTransaction()
tx3a.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0xffffffff),
CTxIn(COutPoint(tx2a_txid, 0), nSequence=0xfffffffd)]
tx3a.vout = [CTxOut(int(0.9*COIN), CScript([b'c'])), CTxOut(int(0.9*COIN), CScript([b'd']))]
tx3a_hex = txToHex(tx3a)
self.nodes[0].sendrawtransaction(tx3a_hex, True)
tx3b = CTransaction()
tx3b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
tx3b.vout = [CTxOut(int(0.5*COIN), CScript([b'e']))]
tx3b_hex = txToHex(tx3b)
tx3c = CTransaction()
tx3c.vin = [CTxIn(COutPoint(tx2a_txid, 0), nSequence=0)]
tx3c.vout = [CTxOut(int(0.5*COIN), CScript([b'f']))]
tx3c_hex = txToHex(tx3c)
self.nodes[0].sendrawtransaction(tx3b_hex, True)
# If tx3b was accepted, tx3c won't look like a replacement,
# but make sure it is accepted anyway
self.nodes[0].sendrawtransaction(tx3c_hex, True)
def test_prioritised_transactions(self):
# Ensure that fee deltas used via prioritisetransaction are
# correctly used by replacement logic
# 1. Check that feeperkb uses modified fees
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Higher fee, but the actual fee per KB is much lower.
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.001*COIN), CScript([b'a'*740000]))]
tx1b_hex = txToHex(tx1b)
# Verify tx1b cannot replace tx1a.
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
# Use prioritisetransaction to set tx1a's fee to 0.
self.nodes[0].prioritisetransaction(txid=tx1a_txid, fee_delta=int(-0.1*COIN))
# Now tx1b should be able to replace tx1a
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
assert(tx1b_txid in self.nodes[0].getrawmempool())
# 2. Check that absolute fee checks use modified fee.
tx1_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2a.vout = [CTxOut(1*COIN, CScript([b'a']))]
tx2a_hex = txToHex(tx2a)
self.nodes[0].sendrawtransaction(tx2a_hex, True)
# Lower fee, but we'll prioritise it
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(1.01*COIN), CScript([b'a']))]
tx2b.rehash()
tx2b_hex = txToHex(tx2b)
# Verify tx2b cannot replace tx2a.
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx2b_hex, True)
# Now prioritise tx2b to have a higher modified fee
self.nodes[0].prioritisetransaction(txid=tx2b.hash, fee_delta=int(0.1*COIN))
# tx2b should now be accepted
tx2b_txid = self.nodes[0].sendrawtransaction(tx2b_hex, True)
assert(tx2b_txid in self.nodes[0].getrawmempool())
def test_rpc(self):
us0 = self.nodes[0].listunspent()[0]
ins = [us0]
outs = {self.nodes[0].getnewaddress() : Decimal(1.0000000)}
rawtx0 = self.nodes[0].createrawtransaction(ins, outs, 0, True)
rawtx1 = self.nodes[0].createrawtransaction(ins, outs, 0, False)
json0 = self.nodes[0].decoderawtransaction(rawtx0)
json1 = self.nodes[0].decoderawtransaction(rawtx1)
assert_equal(json0["vin"][0]["sequence"], 4294967293)
assert_equal(json1["vin"][0]["sequence"], 4294967295)
rawtx2 = self.nodes[0].createrawtransaction([], outs)
frawtx2a = self.nodes[0].fundrawtransaction(rawtx2, {"replaceable": True})
frawtx2b = self.nodes[0].fundrawtransaction(rawtx2, {"replaceable": False})
json0 = self.nodes[0].decoderawtransaction(frawtx2a['hex'])
json1 = self.nodes[0].decoderawtransaction(frawtx2b['hex'])
assert_equal(json0["vin"][0]["sequence"], 4294967293)
assert_equal(json1["vin"][0]["sequence"], 4294967294)
if __name__ == '__main__':
ReplaceByFeeTest().main()