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pytest: simplify test_onchain_multihtlc_our_unilateral/their_unilateral

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We don't actually care how it does it, just that it ends the HTLCs,
so simplify this logic which tries to match it exactly.

This also fixes the flake where we would sometimes close the upstream
channels, simply because we now wait at least one second per block.

Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
This commit is contained in:
Rusty Russell 2022-07-30 09:12:46 +09:30 • committed by neil saitug
parent 66d8ce7c8c
commit b5ee5e7fb1
1 changed files with 19 additions and 86 deletions
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105
tests/test_closing.py
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@ -2861,50 +2861,20 @@ def test_onchain_multihtlc_our_unilateral(node_factory, bitcoind):
l7.daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
l7.daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
# After at depth 5, midnode will spend its own to-self output.
bitcoind.generate_block(4)
l4.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
'OUR_UNILATERAL/DELAYED_OUTPUT_TO_US')
# Rather than track exact complex logic here, simply mine a block
# until l4 says 'all outputs resolved'.
while not l4.daemon.is_in_log('All outputs resolved'):
bitcoind.generate_block(1)
assert bitcoind.rpc.getblockcount() < 200
sync_blockheight(bitcoind, [l4, l5])
# The three outgoing HTLCs time out at 21, 21 and 22 blocks.
bitcoind.generate_block(16)
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
'OUR_UNILATERAL/OUR_HTLC')
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
'OUR_UNILATERAL/OUR_HTLC')
bitcoind.generate_block(1)
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
'OUR_UNILATERAL/OUR_HTLC')
# And three more for us to consider them all settled.
bitcoind.generate_block(3)
# Now, those nodes should have correctly failed the HTLCs
for n in l1, l2:
# All payments should be long resolved.
for n in l1, l2, l3, l5, l6, l7:
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
n.rpc.waitsendpay(h, TIMEOUT)
# Other timeouts are 27,27,28 blocks.
bitcoind.generate_block(2)
l4.daemon.wait_for_logs(['Ignoring output.*: OUR_UNILATERAL/THEIR_HTLC'] * 2)
for _ in range(2):
l5.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
'THEIR_UNILATERAL/OUR_HTLC')
bitcoind.generate_block(1)
l4.daemon.wait_for_log('Ignoring output.*: OUR_UNILATERAL/THEIR_HTLC')
l5.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
'THEIR_UNILATERAL/OUR_HTLC')
# Depth 3 to consider it settled.
bitcoind.generate_block(3)
for n in l5, l6, l7:
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
n.rpc.waitsendpay(h, TIMEOUT)
# At depth 100 it's all done (we didn't bother waiting for mid+1's
# spends, so that might still be going)
bitcoind.generate_block(97)
# At depth 100 it's all done
bitcoind.generate_block(100)
l4.daemon.wait_for_logs(['onchaind complete, forgetting peer'])
# No other channels should have failed.
@ -2947,56 +2917,19 @@ def test_onchain_multihtlc_their_unilateral(node_factory, bitcoind):
l7.daemon.wait_for_log('peer_out WIRE_COMMITMENT_SIGNED')
l7.daemon.wait_for_log('peer_out WIRE_REVOKE_AND_ACK')
# At depth 5, midnode+1 will spend its own to-self output.
bitcoind.generate_block(4)
l5.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET')
# Rather than track exact complex logic here, simply mine a block
# until l5 says 'all outputs resolved'.
while not l5.daemon.is_in_log('All outputs resolved'):
bitcoind.generate_block(1)
assert bitcoind.rpc.getblockcount() < 200
sync_blockheight(bitcoind, [l4, l5])
# The three outgoing HTLCs time out at depth 21, 21 and 22 blocks.
bitcoind.generate_block(16)
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
'THEIR_UNILATERAL/OUR_HTLC')
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
'THEIR_UNILATERAL/OUR_HTLC')
bitcoind.generate_block(1)
l4.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TO_US',
'THEIR_UNILATERAL/OUR_HTLC')
# At depth 3 we consider them all settled.
bitcoind.generate_block(3)
# Now, those nodes should have correctly failed the HTLCs
for n in l1, l2:
# All payments should be long resolved.
for n in l1, l2, l3, l5, l6, l7:
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
n.rpc.waitsendpay(h, TIMEOUT)
# Other timeouts are at depths 27,27,28 blocks.
bitcoind.generate_block(2)
l4.daemon.wait_for_logs(['Ignoring output.*: THEIR_UNILATERAL/THEIR_HTLC'] * 2)
for _ in range(2):
l5.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
'OUR_UNILATERAL/OUR_HTLC')
bitcoind.generate_block(1)
l4.daemon.wait_for_log('Ignoring output.*: THEIR_UNILATERAL/THEIR_HTLC')
l5.wait_for_onchaind_broadcast('OUR_HTLC_TIMEOUT_TX',
'OUR_UNILATERAL/OUR_HTLC')
# At depth 3 we consider them all settled.
bitcoind.generate_block(3)
for n in l5, l6, l7:
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE'):
n.rpc.waitsendpay(h, TIMEOUT)
# At depth 5, mid+1 can spend HTLC_TIMEOUT_TX output.
bitcoind.generate_block(1)
for _ in range(2):
l5.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
bitcoind.generate_block(1)
l5.wait_for_onchaind_broadcast('OUR_DELAYED_RETURN_TO_WALLET',
'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
# At depth 100 they're all done.
# At depth 100 it's all done
bitcoind.generate_block(100)
l4.daemon.wait_for_logs(['onchaind complete, forgetting peer'])
l5.daemon.wait_for_logs(['onchaind complete, forgetting peer'])