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e1dbc0b12b
We reserve inputs when we're going to send a transaction, but we don't unreserve them if we crash. This is most graphically demonstrated by the txprepare case, which makes it easier to trigger. Instead, we should query bitcoind to see whether the tx made it out or not, as we would do manually with dev-rescan-outputs. Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
379 lines
15 KiB
Python
379 lines
15 KiB
Python
from decimal import Decimal
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from fixtures import * # noqa: F401,F403
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from flaky import flaky # noqa: F401
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from lightning import RpcError, Millisatoshi
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from utils import only_one, wait_for
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import pytest
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import time
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def test_withdraw(node_factory, bitcoind):
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amount = 1000000
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# Don't get any funds from previous runs.
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l1 = node_factory.get_node(random_hsm=True)
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l2 = node_factory.get_node(random_hsm=True)
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addr = l1.rpc.newaddr()['bech32']
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# Add some funds to withdraw later
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for i in range(10):
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l1.bitcoin.rpc.sendtoaddress(addr, amount / 10**8 + 0.01)
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 10)
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# Reach around into the db to check that outputs were added
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=0')[0]['c'] == 10
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waddr = l1.bitcoin.rpc.getnewaddress()
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# Now attempt to withdraw some (making sure we collect multiple inputs)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('not an address', amount)
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with pytest.raises(RpcError):
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l1.rpc.withdraw(waddr, 'not an amount')
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with pytest.raises(RpcError):
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l1.rpc.withdraw(waddr, -amount)
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with pytest.raises(RpcError, match=r'Cannot afford transaction'):
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l1.rpc.withdraw(waddr, amount * 100)
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out = l1.rpc.withdraw(waddr, 2 * amount)
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# Make sure bitcoind received the withdrawal
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unspent = l1.bitcoin.rpc.listunspent(0)
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withdrawal = [u for u in unspent if u['txid'] == out['txid']]
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assert(withdrawal[0]['amount'] == Decimal('0.02'))
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# Now make sure two of them were marked as spent
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=2')[0]['c'] == 2
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# Now send some money to l2.
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# lightningd uses P2SH-P2WPKH
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waddr = l2.rpc.newaddr('bech32')['bech32']
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l1.rpc.withdraw(waddr, 2 * amount)
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bitcoind.generate_block(1)
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# Make sure l2 received the withdrawal.
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wait_for(lambda: len(l2.rpc.listfunds()['outputs']) == 1)
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outputs = l2.db_query('SELECT value FROM outputs WHERE status=0;')
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assert only_one(outputs)['value'] == 2 * amount
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# Now make sure an additional two of them were marked as spent
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=2')[0]['c'] == 4
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# Simple test for withdrawal to P2WPKH
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# Address from: https://bc-2.jp/tools/bech32demo/index.html
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waddr = 'bcrt1qw508d6qejxtdg4y5r3zarvary0c5xw7kygt080'
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with pytest.raises(RpcError):
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l1.rpc.withdraw('xx1qw508d6qejxtdg4y5r3zarvary0c5xw7kxpjzsx', 2 * amount)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1pw508d6qejxtdg4y5r3zarvary0c5xw7kdl9fad', 2 * amount)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1qw508d6qejxtdg4y5r3zarvary0c5xw7kxxxxxx', 2 * amount)
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l1.rpc.withdraw(waddr, 2 * amount)
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bitcoind.generate_block(1)
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# Now make sure additional two of them were marked as spent
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=2')[0]['c'] == 6
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# Simple test for withdrawal to P2WSH
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# Address from: https://bc-2.jp/tools/bech32demo/index.html
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waddr = 'bcrt1qrp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3qzf4jry'
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with pytest.raises(RpcError):
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l1.rpc.withdraw('xx1qrp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3q0sl5k7', 2 * amount)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1prp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3qsm03tq', 2 * amount)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1qrp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3qxxxxxx', 2 * amount)
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l1.rpc.withdraw(waddr, 2 * amount)
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bitcoind.generate_block(1)
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# Now make sure additional two of them were marked as spent
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=2')[0]['c'] == 8
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# failure testing for invalid SegWit addresses, from BIP173
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# HRP character out of range
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with pytest.raises(RpcError):
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l1.rpc.withdraw(' 1nwldj5', 2 * amount)
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# overall max length exceeded
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with pytest.raises(RpcError):
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l1.rpc.withdraw('an84characterslonghumanreadablepartthatcontainsthenumber1andtheexcludedcharactersbio1569pvx', 2 * amount)
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# No separator character
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with pytest.raises(RpcError):
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l1.rpc.withdraw('pzry9x0s0muk', 2 * amount)
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# Empty HRP
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with pytest.raises(RpcError):
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l1.rpc.withdraw('1pzry9x0s0muk', 2 * amount)
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# Invalid witness version
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with pytest.raises(RpcError):
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l1.rpc.withdraw('BC13W508D6QEJXTDG4Y5R3ZARVARY0C5XW7KN40WF2', 2 * amount)
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# Invalid program length for witness version 0 (per BIP141)
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with pytest.raises(RpcError):
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l1.rpc.withdraw('BC1QR508D6QEJXTDG4Y5R3ZARVARYV98GJ9P', 2 * amount)
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# Mixed case
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1qrp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3q0sL5k7', 2 * amount)
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# Non-zero padding in 8-to-5 conversion
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with pytest.raises(RpcError):
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l1.rpc.withdraw('tb1qrp33g0q5c5txsp9arysrx4k6zdkfs4nce4xj0gdcccefvpysxf3pjxtptv', 2 * amount)
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# Should have 6 outputs available.
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=0')[0]['c'] == 6
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# Test withdrawal to self.
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l1.rpc.withdraw(l1.rpc.newaddr('bech32')['bech32'], 'all', minconf=0)
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bitcoind.generate_block(1)
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=0')[0]['c'] == 1
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l1.rpc.withdraw(waddr, 'all', minconf=0)
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assert l1.db_query('SELECT COUNT(*) as c FROM outputs WHERE status=0')[0]['c'] == 0
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# This should fail, can't even afford fee.
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with pytest.raises(RpcError, match=r'Cannot afford transaction'):
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l1.rpc.withdraw(waddr, 'all')
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def test_minconf_withdraw(node_factory, bitcoind):
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"""Issue 2518: ensure that ridiculous confirmation levels don't overflow
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The number of confirmations is used to compute a maximum height that is to
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be accepted. If the current height is smaller than the number of
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confirmations we wrap around and just select everything. The fix is to
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clamp the maxheight parameter to a positive small number.
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"""
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amount = 1000000
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# Don't get any funds from previous runs.
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l1 = node_factory.get_node(random_hsm=True)
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addr = l1.rpc.newaddr()['bech32']
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# Add some funds to withdraw later
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for i in range(10):
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l1.bitcoin.rpc.sendtoaddress(addr, amount / 10**8 + 0.01)
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 10)
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with pytest.raises(RpcError):
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l1.rpc.withdraw(destination=addr, satoshi=10000, feerate='normal', minconf=9999999)
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def test_addfunds_from_block(node_factory, bitcoind):
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"""Send funds to the daemon without telling it explicitly
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"""
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# Previous runs with same bitcoind can leave funds!
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l1 = node_factory.get_node(random_hsm=True)
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addr = l1.rpc.newaddr()['bech32']
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bitcoind.rpc.sendtoaddress(addr, 0.1)
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 1)
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outputs = l1.db_query('SELECT value FROM outputs WHERE status=0;')
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assert only_one(outputs)['value'] == 10000000
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# The address we detect must match what was paid to.
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output = only_one(l1.rpc.listfunds()['outputs'])
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assert output['address'] == addr
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# Send all our money to a P2WPKH address this time.
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addr = l1.rpc.newaddr("bech32")['bech32']
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l1.rpc.withdraw(addr, "all")
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bitcoind.generate_block(1)
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time.sleep(1)
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# The address we detect must match what was paid to.
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output = only_one(l1.rpc.listfunds()['outputs'])
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assert output['address'] == addr
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def test_txprepare(node_factory, bitcoind):
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amount = 1000000
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l1 = node_factory.get_node(random_hsm=True)
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# Add some funds to withdraw later: both bech32 and p2sh
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for i in range(5):
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr()['bech32'],
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amount / 10**8)
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr('p2sh-segwit')['p2sh-segwit'],
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amount / 10**8)
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 10)
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prep = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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Millisatoshi(amount * 3 * 1000))
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decode = bitcoind.rpc.decoderawtransaction(prep['unsigned_tx'])
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assert decode['txid'] == prep['txid']
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# 4 inputs, 2 outputs.
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assert len(decode['vin']) == 4
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assert len(decode['vout']) == 2
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# One output will be correct.
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if decode['vout'][0]['value'] == Decimal(amount * 3) / 10**8:
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outnum = 0
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changenum = 1
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elif decode['vout'][1]['value'] == Decimal(amount * 3) / 10**8:
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outnum = 1
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changenum = 0
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else:
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assert False
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assert decode['vout'][outnum]['scriptPubKey']['type'] == 'witness_v0_keyhash'
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assert decode['vout'][outnum]['scriptPubKey']['addresses'] == ['bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg']
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assert decode['vout'][changenum]['scriptPubKey']['type'] == 'witness_v0_keyhash'
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# Now prepare one with no change.
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prep2 = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep2['unsigned_tx'])
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assert decode['txid'] == prep2['txid']
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# 6 inputs, 1 outputs.
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assert len(decode['vin']) == 6
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assert len(decode['vout']) == 1
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# Some fees will be paid.
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assert decode['vout'][0]['value'] < Decimal(amount * 6) / 10**8
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assert decode['vout'][0]['value'] > Decimal(amount * 6) / 10**8 - Decimal(0.0002)
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assert decode['vout'][0]['scriptPubKey']['type'] == 'witness_v0_keyhash'
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assert decode['vout'][0]['scriptPubKey']['addresses'] == ['bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg']
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# If I cancel the first one, I can get those first 4 outputs.
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discard = l1.rpc.txdiscard(prep['txid'])
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assert discard['txid'] == prep['txid']
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assert discard['unsigned_tx'] == prep['unsigned_tx']
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prep3 = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep3['unsigned_tx'])
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assert decode['txid'] == prep3['txid']
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# 4 inputs, 1 outputs.
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assert len(decode['vin']) == 4
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assert len(decode['vout']) == 1
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# Some fees will be taken
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assert decode['vout'][0]['value'] < Decimal(amount * 4) / 10**8
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assert decode['vout'][0]['value'] > Decimal(amount * 4) / 10**8 - Decimal(0.0002)
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assert decode['vout'][0]['scriptPubKey']['type'] == 'witness_v0_keyhash'
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assert decode['vout'][0]['scriptPubKey']['addresses'] == ['bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg']
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# Cannot discard twice.
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with pytest.raises(RpcError, match=r'not an unreleased txid'):
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l1.rpc.txdiscard(prep['txid'])
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# Discard everything, we should now spend all inputs.
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l1.rpc.txdiscard(prep2['txid'])
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l1.rpc.txdiscard(prep3['txid'])
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prep4 = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep4['unsigned_tx'])
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assert decode['txid'] == prep4['txid']
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# 10 inputs, 1 outputs.
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assert len(decode['vin']) == 10
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assert len(decode['vout']) == 1
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# Some fees will be taken
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assert decode['vout'][0]['value'] < Decimal(amount * 10) / 10**8
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assert decode['vout'][0]['value'] > Decimal(amount * 10) / 10**8 - Decimal(0.0003)
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assert decode['vout'][0]['scriptPubKey']['type'] == 'witness_v0_keyhash'
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assert decode['vout'][0]['scriptPubKey']['addresses'] == ['bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg']
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def test_txsend(node_factory, bitcoind):
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amount = 1000000
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l1 = node_factory.get_node(random_hsm=True)
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# Add some funds to withdraw later: both bech32 and p2sh
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for i in range(5):
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr()['bech32'],
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amount / 10**8)
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr('p2sh-segwit')['p2sh-segwit'],
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amount / 10**8)
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bitcoind.generate_block(1)
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 10)
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prep = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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Millisatoshi(amount * 3 * 1000))
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out = l1.rpc.txsend(prep['txid'])
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# Cannot discard after send!
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with pytest.raises(RpcError, match=r'not an unreleased txid'):
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l1.rpc.txdiscard(prep['txid'])
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wait_for(lambda: prep['txid'] in bitcoind.rpc.getrawmempool())
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# Signed tx should have same txid
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decode = bitcoind.rpc.decoderawtransaction(out['tx'])
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assert decode['txid'] == prep['txid']
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bitcoind.generate_block(1)
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# Change output should appear.
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if decode['vout'][0]['value'] == Decimal(amount * 3) / 10**8:
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changenum = 1
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elif decode['vout'][1]['value'] == Decimal(amount * 3) / 10**8:
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changenum = 0
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else:
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assert False
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# Those spent outputs are gone, but change output has arrived.
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wait_for(lambda: len(l1.rpc.listfunds()['outputs']) == 10 - len(decode['vin']) + 1)
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# Change address should appear in listfunds()
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assert decode['vout'][changenum]['scriptPubKey']['addresses'][0] in [f['address'] for f in l1.rpc.listfunds()['outputs']]
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def test_txprepare_restart(node_factory, bitcoind):
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amount = 1000000
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l1 = node_factory.get_node(may_fail=True)
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# Add some funds to withdraw later: both bech32 and p2sh
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for i in range(5):
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr()['bech32'],
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amount / 10**8)
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bitcoind.rpc.sendtoaddress(l1.rpc.newaddr('p2sh-segwit')['p2sh-segwit'],
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amount / 10**8)
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bitcoind.generate_block(1)
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wait_for(lambda: [o['status'] for o in l1.rpc.listfunds()['outputs']] == ['confirmed'] * 10)
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prep = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep['unsigned_tx'])
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assert decode['txid'] == prep['txid']
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# All 10 inputs
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assert len(decode['vin']) == 10
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# L1 will forget all about it.
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l1.restart()
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# It goes backwards in blockchain just in case there was a reorg. Wait.
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wait_for(lambda: [o['status'] for o in l1.rpc.listfunds()['outputs']] == ['confirmed'] * 10)
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with pytest.raises(RpcError, match=r'not an unreleased txid'):
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l1.rpc.txdiscard(prep['txid'])
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prep = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep['unsigned_tx'])
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assert decode['txid'] == prep['txid']
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# All 10 inputs
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assert len(decode['vin']) == 10
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# This will also work if we simply kill it.
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l1.restart(clean=False)
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# It goes backwards in blockchain just in case there was a reorg. Wait.
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wait_for(lambda: [o['status'] for o in l1.rpc.listfunds()['outputs']] == ['confirmed'] * 10)
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# It should have logged this for each output.
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for i in decode['vin']:
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assert l1.daemon.is_in_log('wallet: reserved output {}/{} reset to available'.format(i['txid'], i['vout']))
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prep = l1.rpc.txprepare('bcrt1qeyyk6sl5pr49ycpqyckvmttus5ttj25pd0zpvg',
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'all')
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decode = bitcoind.rpc.decoderawtransaction(prep['unsigned_tx'])
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assert decode['txid'] == prep['txid']
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# All 10 inputs
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assert len(decode['vin']) == 10
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