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Diffstat (limited to 'test/functional/feature_rbf.py')
| -rwxr-xr-x | test/functional/feature_rbf.py | 565 |
1 files changed, 565 insertions, 0 deletions
diff --git a/test/functional/feature_rbf.py b/test/functional/feature_rbf.py new file mode 100755 index 0000000000..d6ab5ecc37 --- /dev/null +++ b/test/functional/feature_rbf.py @@ -0,0 +1,565 @@ +#!/usr/bin/env python3 +# Copyright (c) 2014-2017 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.signrawtransactionwithwallet(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() |