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Add RPC test for abandoned and conflicted transactions.

pull/1/head
Alex Morcos 7 years ago
parent
commit
df0e2226d9
  1. 1
      qa/pull-tester/rpc-tests.py
  2. 153
      qa/rpc-tests/abandonconflict.py

1
qa/pull-tester/rpc-tests.py

@ -105,6 +105,7 @@ testScripts = [ @@ -105,6 +105,7 @@ testScripts = [
'prioritise_transaction.py',
'invalidblockrequest.py',
'invalidtxrequest.py',
'abandonconflict.py',
]
testScriptsExt = [
'bip65-cltv.py',

153
qa/rpc-tests/abandonconflict.py

@ -0,0 +1,153 @@ @@ -0,0 +1,153 @@
#!/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"))
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()
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