#!/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) tx1a_txid = 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) tx1_txid = 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) tx2a_txid = 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()