Python CTransaction.vin方法代码示例

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
        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' * 35]))]
        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' * 35]))]
        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())

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
            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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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()

        feeout = CTxOut(int(0.1*COIN), CScript())
        tx1a = CTransaction()
        tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
        tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35])), feeout]
        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' * 35])), feeout]
        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' * 35])), feeout, feeout]
        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

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
        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

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
        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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
        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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    def test_bip68_not_consensus(self):
        assert(get_bip9_status(self.nodes[0], 'csv')['status'] != 'active')
        txid = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), 2)

        tx1 = FromHex(CTransaction(), self.nodes[0].getrawtransaction(txid))
        tx1.rehash()

        # Make an anyone-can-spend transaction
        tx2 = CTransaction()
        tx2.nVersion = 1
        tx2.vin = [CTxIn(COutPoint(tx1.sha256, 0), nSequence=0)]
        tx2.vout = [CTxOut(int(tx1.vout[0].nValue - self.relayfee*COIN), CScript([b'a']))]

        # sign tx2
        tx2_raw = self.nodes[0].signrawtransactionwithwallet(ToHex(tx2))["hex"]
        tx2 = FromHex(tx2, tx2_raw)
        tx2.rehash()

        self.nodes[0].sendrawtransaction(ToHex(tx2))

        # Now make an invalid spend of tx2 according to BIP68
        sequence_value = 100 # 100 block relative locktime

        tx3 = CTransaction()
        tx3.nVersion = 2
        tx3.vin = [CTxIn(COutPoint(tx2.sha256, 0), nSequence=sequence_value)]
        tx3.vout = [CTxOut(int(tx2.vout[0].nValue - self.relayfee * COIN), CScript([b'a' * 35]))]
        tx3.rehash()

        assert_raises_rpc_error(-26, NOT_FINAL_ERROR, self.nodes[0].sendrawtransaction, ToHex(tx3))

        # make a block that violates bip68; ensure that the tip updates
        tip = int(self.nodes[0].getbestblockhash(), 16)
        block = create_block(tip, create_coinbase(self.nodes[0].getblockcount()+1))
        block.nVersion = 3
        block.vtx.extend([tx1, tx2, tx3])
        block.hashMerkleRoot = block.calc_merkle_root()
        block.rehash()
        add_witness_commitment(block)
        block.solve()

        self.nodes[0].submitblock(bytes_to_hex_str(block.serialize(True)))
        assert_equal(self.nodes[0].getbestblockhash(), block.hash)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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)]
            feeout = CTxOut(1*COIN)
            tx.vout = [CTxOut(remaining_value, CScript([1, OP_DROP] * 15 + [1])), feeout]
            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] * 35)), CTxOut(30*COIN)]
        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] * 35)), CTxOut(49*COIN)]
        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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    def test_disable_flag(self):
        # Create some unconfirmed inputs
        new_addr = self.nodes[0].getnewaddress()
        self.nodes[0].sendtoaddress(new_addr, 2) # send 2 BTC

        utxos = self.nodes[0].listunspent(0, 0)
        assert len(utxos) > 0

        utxo = utxos[0]

        tx1 = CTransaction()
        value = int(satoshi_round(utxo["amount"] - self.relayfee)*COIN)

        # Check that the disable flag disables relative locktime.
        # If sequence locks were used, this would require 1 block for the
        # input to mature.
        sequence_value = SEQUENCE_LOCKTIME_DISABLE_FLAG | 1
        tx1.vin = [CTxIn(COutPoint(int(utxo["txid"], 16), utxo["vout"]), nSequence=sequence_value)]
        tx1.vout = [CTxOut(value, CScript([b'a']))]

        tx1_signed = self.nodes[0].signrawtransactionwithwallet(ToHex(tx1))["hex"]
        tx1_id = self.nodes[0].sendrawtransaction(tx1_signed)
        tx1_id = int(tx1_id, 16)

        # This transaction will enable sequence-locks, so this transaction should
        # fail
        tx2 = CTransaction()
        tx2.nVersion = 2
        sequence_value = sequence_value & 0x7fffffff
        tx2.vin = [CTxIn(COutPoint(tx1_id, 0), nSequence=sequence_value)]
        tx2.vout = [CTxOut(int(value - self.relayfee * COIN), CScript([b'a' * 35]))]
        tx2.rehash()

        assert_raises_rpc_error(-26, NOT_FINAL_ERROR, self.nodes[0].sendrawtransaction, ToHex(tx2))

        # Setting the version back down to 1 should disable the sequence lock,
        # so this should be accepted.
        tx2.nVersion = 1

        self.nodes[0].sendrawtransaction(ToHex(tx2))

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
        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' * 35]))]
        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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
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()['bitcoin'] < satoshi_round((amount + fee)/COIN):
        node.generate(100)

    new_addr = node.getnewaddress()
    unblinded_addr = node.validateaddress(new_addr)["unconfidential"]
    txidstr = node.sendtoaddress(new_addr, satoshi_round((amount+fee)/COIN))
    tx1 = node.getrawtransaction(txidstr, 1)
    txid = int(txidstr, 16)
    i = None

    for i, txout in enumerate(tx1['vout']):
        if txout['scriptPubKey']['type'] == "fee":
            continue # skip fee outputs
        if txout['scriptPubKey']['addresses'] == [unblinded_addr]:
            break
    assert i is not None

    tx2 = CTransaction()
    tx2.vin = [CTxIn(COutPoint(txid, i))]
    tx1raw = CTransaction()
    tx1raw.deserialize(BytesIO(hex_str_to_bytes(node.getrawtransaction(txidstr))))
    feeout = CTxOut(CTxOutValue(tx1raw.vout[i].nValue.getAmount() - amount))
    tx2.vout = [CTxOut(amount, scriptPubKey), feeout]
    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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
        def test_nonzero_locks(orig_tx, node, relayfee, use_height_lock):
            sequence_value = 1
            if not use_height_lock:
                sequence_value |= SEQUENCE_LOCKTIME_TYPE_FLAG

            tx = CTransaction()
            tx.nVersion = 2
            tx.vin = [CTxIn(COutPoint(orig_tx.sha256, 0), nSequence=sequence_value)]
            tx.vout = [CTxOut(int(orig_tx.vout[0].nValue - relayfee * COIN), CScript([b'a' * 35]))]
            tx.rehash()

            if (orig_tx.hash in node.getrawmempool()):
                # sendrawtransaction should fail if the tx is in the mempool
                assert_raises_rpc_error(-26, NOT_FINAL_ERROR, node.sendrawtransaction, ToHex(tx))
            else:
                # sendrawtransaction should succeed if the tx is not in the mempool
                node.sendrawtransaction(ToHex(tx))

            return tx

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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' * 35]))]
        tx1a_hex = txToHex(tx1a)
        tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)

        # This transaction isn't shown as replaceable
        assert_equal(self.nodes[0].getmempoolentry(tx1a_txid)['bip125-replaceable'], False)

        # 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' * 35]))]
        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' * 35]))]
        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' * 35]))]
        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)

        tx3a_txid = self.nodes[0].sendrawtransaction(tx3a_hex, True)

        # This transaction is shown as replaceable
        assert_equal(self.nodes[0].getmempoolentry(tx3a_txid)['bip125-replaceable'], True)

        tx3b = CTransaction()
        tx3b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
        tx3b.vout = [CTxOut(int(0.5 * COIN), CScript([b'e' * 35]))]
        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' * 35]))]
        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)

# 需要导入模块: from test_framework.messages import CTransaction [as 别名]
# 或者: from test_framework.messages.CTransaction import vin [as 别名]
    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] * 35))]
        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] * 35))]
        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] * 35))]
            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)