kevacoin/qa/rpc-tests/abandonconflict.py

#!/usr/bin/env python2
# Copyright (c) 2014-2015 The Bitcoin Core developers
# Distributed under the MIT software license, see the accompanying
# file COPYING or http://www.opensource.org/licenses/mit-license.php.


from test_framework.test_framework import BitcoinTestFramework
from test_framework.util import *
try:
    import urllib.parse as urlparse
except ImportError:
    import urlparse

class AbandonConflictTest(BitcoinTestFramework):

    def setup_network(self):
        self.nodes = []
        self.nodes.append(start_node(0, self.options.tmpdir, ["-debug","-logtimemicros","-minrelaytxfee=0.00001"]))
        self.nodes.append(start_node(1, self.options.tmpdir, ["-debug","-logtimemicros"]))
        connect_nodes(self.nodes[0], 1)

    def run_test(self):
        self.nodes[1].generate(100)
        sync_blocks(self.nodes)
        balance = self.nodes[0].getbalance()
        txA = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
        txB = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
        txC = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
        sync_mempools(self.nodes)
        self.nodes[1].generate(1)

        sync_blocks(self.nodes)
        newbalance = self.nodes[0].getbalance()
        assert(balance - newbalance < Decimal("0.001")) #no more than fees lost
        balance = newbalance

        url = urlparse.urlparse(self.nodes[1].url)
        self.nodes[0].disconnectnode(url.hostname+":"+str(p2p_port(1)))

        # Identify the 10btc outputs
        nA = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txA, 1)["vout"]) if vout["value"] == Decimal("10"))
        nB = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txB, 1)["vout"]) if vout["value"] == Decimal("10"))
        nC = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txC, 1)["vout"]) if vout["value"] == Decimal("10"))

        inputs =[]
        # spend 10btc outputs from txA and txB
        inputs.append({"txid":txA, "vout":nA})
        inputs.append({"txid":txB, "vout":nB})
        outputs = {}

        outputs[self.nodes[0].getnewaddress()] = Decimal("14.99998")
        outputs[self.nodes[1].getnewaddress()] = Decimal("5")
        signed = self.nodes[0].signrawtransaction(self.nodes[0].createrawtransaction(inputs, outputs))
        txAB1 = self.nodes[0].sendrawtransaction(signed["hex"])

        # Identify the 14.99998btc output
        nAB = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txAB1, 1)["vout"]) if vout["value"] == Decimal("14.99998"))

        #Create a child tx spending AB1 and C
        inputs = []
        inputs.append({"txid":txAB1, "vout":nAB})
        inputs.append({"txid":txC, "vout":nC})
        outputs = {}
        outputs[self.nodes[0].getnewaddress()] = Decimal("24.9996")
        signed2 = self.nodes[0].signrawtransaction(self.nodes[0].createrawtransaction(inputs, outputs))
        txABC2 = self.nodes[0].sendrawtransaction(signed2["hex"])

        # In mempool txs from self should increase balance from change
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance - Decimal("30") + Decimal("24.9996"))
        balance = newbalance

        # Restart the node with a higher min relay fee so the parent tx is no longer in mempool
        # TODO: redo with eviction
        # Note had to make sure tx did not have AllowFree priority
        stop_node(self.nodes[0],0)
        self.nodes[0]=start_node(0, self.options.tmpdir, ["-debug","-logtimemicros","-minrelaytxfee=0.0001"])

        # Verify txs no longer in mempool
        assert(len(self.nodes[0].getrawmempool()) == 0)

        # Not in mempool txs from self should only reduce balance
        # inputs are still spent, but change not received
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance - Decimal("24.9996"))
        # Unconfirmed received funds that are not in mempool, also shouldn't show
        # up in unconfirmed balance
        unconfbalance = self.nodes[0].getunconfirmedbalance() + self.nodes[0].getbalance()
        assert(unconfbalance == newbalance)
        # Also shouldn't show up in listunspent
        assert(not txABC2 in [utxo["txid"] for utxo in self.nodes[0].listunspent(0)])
        balance = newbalance

        # Abandon original transaction and verify inputs are available again
        # including that the child tx was also abandoned
        self.nodes[0].abandontransaction(txAB1)
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance + Decimal("30"))
        balance = newbalance

        # Verify that even with a low min relay fee, the tx is not reaccepted from wallet on startup once abandoned
        stop_node(self.nodes[0],0)
        self.nodes[0]=start_node(0, self.options.tmpdir, ["-debug","-logtimemicros","-minrelaytxfee=0.00001"])
        assert(len(self.nodes[0].getrawmempool()) == 0)
        assert(self.nodes[0].getbalance() == balance)

        # But if its received again then it is unabandoned
        # And since now in mempool, the change is available
        # But its child tx remains abandoned
        self.nodes[0].sendrawtransaction(signed["hex"])
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance - Decimal("20") + Decimal("14.99998"))
        balance = newbalance

        # Send child tx again so its unabandoned
        self.nodes[0].sendrawtransaction(signed2["hex"])
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance - Decimal("10") - Decimal("14.99998") + Decimal("24.9996"))
        balance = newbalance

        # Remove using high relay fee again
        stop_node(self.nodes[0],0)
        self.nodes[0]=start_node(0, self.options.tmpdir, ["-debug","-logtimemicros","-minrelaytxfee=0.0001"])
        assert(len(self.nodes[0].getrawmempool()) == 0)
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance - Decimal("24.9996"))
        balance = newbalance

        # Create a double spend of AB1 by spending again from only A's 10 output
        # Mine double spend from node 1
        inputs =[]
        inputs.append({"txid":txA, "vout":nA})
        outputs = {}
        outputs[self.nodes[1].getnewaddress()] = Decimal("9.9999")
        tx = self.nodes[0].createrawtransaction(inputs, outputs)
        signed = self.nodes[0].signrawtransaction(tx)
        self.nodes[1].sendrawtransaction(signed["hex"])
        self.nodes[1].generate(1)

        connect_nodes(self.nodes[0], 1)
        sync_blocks(self.nodes)

        # Verify that B and C's 10 BTC outputs are available for spending again because AB1 is now conflicted
        newbalance = self.nodes[0].getbalance()
        assert(newbalance == balance + Decimal("20"))
        balance = newbalance

        # There is currently a minor bug around this and so this test doesn't work.  See Issue #7315
        # Invalidate the block with the double spend and B's 10 BTC output should no longer be available
        # Don't think C's should either
        self.nodes[0].invalidateblock(self.nodes[0].getbestblockhash())
        newbalance = self.nodes[0].getbalance()
        #assert(newbalance == balance - Decimal("10"))
        print "If balance has not declined after invalidateblock then out of mempool wallet tx which is no longer"
        print "conflicted has not resumed causing its inputs to be seen as spent.  See Issue #7315"
        print balance , " -> " , newbalance , " ?"

if __name__ == '__main__':
    AbandonConflictTest().main()