89e70f9d7f
Prior to this change, it would mark only the first layer of child transactions abandoned, due to always following the input hashTx rather than the current now tx.
172 lines
8.3 KiB
Python
Executable file
172 lines
8.3 KiB
Python
Executable file
#!/usr/bin/env python3
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# Copyright (c) 2014-2017 The Bitcoin Core developers
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# Distributed under the MIT software license, see the accompanying
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# file COPYING or http://www.opensource.org/licenses/mit-license.php.
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"""Test the abandontransaction RPC.
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The abandontransaction RPC marks a transaction and all its in-wallet
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descendants as abandoned which allows their inputs to be respent. It can be
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used to replace "stuck" or evicted transactions. It only works on transactions
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which are not included in a block and are not currently in the mempool. It has
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no effect on transactions which are already abandoned.
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"""
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from test_framework.test_framework import BitcoinTestFramework
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from test_framework.util import *
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class AbandonConflictTest(BitcoinTestFramework):
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def set_test_params(self):
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self.num_nodes = 2
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self.extra_args = [["-minrelaytxfee=0.00001"], []]
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def run_test(self):
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self.nodes[1].generate(100)
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sync_blocks(self.nodes)
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balance = self.nodes[0].getbalance()
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txA = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
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txB = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
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txC = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), Decimal("10"))
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sync_mempools(self.nodes)
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self.nodes[1].generate(1)
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# Can not abandon non-wallet transaction
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assert_raises_rpc_error(-5, 'Invalid or non-wallet transaction id', lambda: self.nodes[0].abandontransaction(txid='ff' * 32))
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# Can not abandon confirmed transaction
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assert_raises_rpc_error(-5, 'Transaction not eligible for abandonment', lambda: self.nodes[0].abandontransaction(txid=txA))
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sync_blocks(self.nodes)
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newbalance = self.nodes[0].getbalance()
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assert(balance - newbalance < Decimal("0.001")) #no more than fees lost
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balance = newbalance
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# Disconnect nodes so node0's transactions don't get into node1's mempool
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disconnect_nodes(self.nodes[0], 1)
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# Identify the 10btc outputs
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nA = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txA, 1)["vout"]) if vout["value"] == Decimal("10"))
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nB = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txB, 1)["vout"]) if vout["value"] == Decimal("10"))
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nC = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txC, 1)["vout"]) if vout["value"] == Decimal("10"))
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inputs =[]
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# spend 10btc outputs from txA and txB
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inputs.append({"txid":txA, "vout":nA})
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inputs.append({"txid":txB, "vout":nB})
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outputs = {}
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outputs[self.nodes[0].getnewaddress()] = Decimal("14.99998")
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outputs[self.nodes[1].getnewaddress()] = Decimal("5")
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signed = self.nodes[0].signrawtransactionwithwallet(self.nodes[0].createrawtransaction(inputs, outputs))
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txAB1 = self.nodes[0].sendrawtransaction(signed["hex"])
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# Identify the 14.99998btc output
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nAB = next(i for i, vout in enumerate(self.nodes[0].getrawtransaction(txAB1, 1)["vout"]) if vout["value"] == Decimal("14.99998"))
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#Create a child tx spending AB1 and C
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inputs = []
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inputs.append({"txid":txAB1, "vout":nAB})
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inputs.append({"txid":txC, "vout":nC})
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outputs = {}
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outputs[self.nodes[0].getnewaddress()] = Decimal("24.9996")
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signed2 = self.nodes[0].signrawtransactionwithwallet(self.nodes[0].createrawtransaction(inputs, outputs))
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txABC2 = self.nodes[0].sendrawtransaction(signed2["hex"])
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# Create a child tx spending ABC2
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signed3_change = Decimal("24.999")
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inputs = [ {"txid":txABC2, "vout":0} ]
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outputs = { self.nodes[0].getnewaddress(): signed3_change }
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signed3 = self.nodes[0].signrawtransactionwithwallet(self.nodes[0].createrawtransaction(inputs, outputs))
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# note tx is never directly referenced, only abandoned as a child of the above
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self.nodes[0].sendrawtransaction(signed3["hex"])
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# In mempool txs from self should increase balance from change
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance - Decimal("30") + signed3_change)
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balance = newbalance
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# Restart the node with a higher min relay fee so the parent tx is no longer in mempool
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# TODO: redo with eviction
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self.stop_node(0)
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self.start_node(0, extra_args=["-minrelaytxfee=0.0001"])
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# Verify txs no longer in either node's mempool
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assert_equal(len(self.nodes[0].getrawmempool()), 0)
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assert_equal(len(self.nodes[1].getrawmempool()), 0)
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# Not in mempool txs from self should only reduce balance
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# inputs are still spent, but change not received
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance - signed3_change)
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# Unconfirmed received funds that are not in mempool, also shouldn't show
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# up in unconfirmed balance
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unconfbalance = self.nodes[0].getunconfirmedbalance() + self.nodes[0].getbalance()
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assert_equal(unconfbalance, newbalance)
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# Also shouldn't show up in listunspent
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assert(not txABC2 in [utxo["txid"] for utxo in self.nodes[0].listunspent(0)])
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balance = newbalance
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# Abandon original transaction and verify inputs are available again
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# including that the child tx was also abandoned
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self.nodes[0].abandontransaction(txAB1)
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance + Decimal("30"))
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balance = newbalance
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# Verify that even with a low min relay fee, the tx is not reaccepted from wallet on startup once abandoned
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self.stop_node(0)
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self.start_node(0, extra_args=["-minrelaytxfee=0.00001"])
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assert_equal(len(self.nodes[0].getrawmempool()), 0)
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assert_equal(self.nodes[0].getbalance(), balance)
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# But if it is received again then it is unabandoned
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# And since now in mempool, the change is available
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# But its child tx remains abandoned
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self.nodes[0].sendrawtransaction(signed["hex"])
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance - Decimal("20") + Decimal("14.99998"))
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balance = newbalance
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# Send child tx again so it is unabandoned
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self.nodes[0].sendrawtransaction(signed2["hex"])
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance - Decimal("10") - Decimal("14.99998") + Decimal("24.9996"))
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balance = newbalance
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# Remove using high relay fee again
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self.stop_node(0)
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self.start_node(0, extra_args=["-minrelaytxfee=0.0001"])
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assert_equal(len(self.nodes[0].getrawmempool()), 0)
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance - Decimal("24.9996"))
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balance = newbalance
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# Create a double spend of AB1 by spending again from only A's 10 output
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# Mine double spend from node 1
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inputs =[]
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inputs.append({"txid":txA, "vout":nA})
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outputs = {}
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outputs[self.nodes[1].getnewaddress()] = Decimal("9.9999")
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tx = self.nodes[0].createrawtransaction(inputs, outputs)
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signed = self.nodes[0].signrawtransactionwithwallet(tx)
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self.nodes[1].sendrawtransaction(signed["hex"])
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self.nodes[1].generate(1)
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connect_nodes(self.nodes[0], 1)
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sync_blocks(self.nodes)
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# Verify that B and C's 10 BTC outputs are available for spending again because AB1 is now conflicted
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newbalance = self.nodes[0].getbalance()
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assert_equal(newbalance, balance + Decimal("20"))
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balance = newbalance
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# There is currently a minor bug around this and so this test doesn't work. See Issue #7315
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# Invalidate the block with the double spend and B's 10 BTC output should no longer be available
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# Don't think C's should either
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self.nodes[0].invalidateblock(self.nodes[0].getbestblockhash())
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newbalance = self.nodes[0].getbalance()
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#assert_equal(newbalance, balance - Decimal("10"))
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self.log.info("If balance has not declined after invalidateblock then out of mempool wallet tx which is no longer")
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self.log.info("conflicted has not resumed causing its inputs to be seen as spent. See Issue #7315")
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self.log.info(str(balance) + " -> " + str(newbalance) + " ?")
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if __name__ == '__main__':
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AbandonConflictTest().main()
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