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pytest: make test_onchain_close_upstream more accurately reflect report. 

1. Don't use dust HTLCs.
2. Make l3 unresponsive, like report.
3. Make l2-l3 fail because we time out on successive HTLC.

We use sendpay rather than pay, because pay can do multiple attempts.

Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
This commit is contained in:
Rusty Russell 2022-03-30 14:13:08 +10:30
parent 1e08c3b882
commit 5033e22835
1 changed files with 40 additions and 10 deletions
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50
tests/test_closing.py
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@ -3634,22 +3634,45 @@ def test_close_weight_estimate(node_factory, bitcoind):
@pytest.mark.developer("needs dev_disconnect")
def test_onchain_close_upstream(node_factory, bitcoind, executor):
def test_onchain_close_upstream(node_factory, bitcoind):
"""https://github.com/ElementsProject/lightning/issues/4649
We send an HTLC, and peer unilaterally closes: do we close upstream?
"""
l1, l2, l3 = node_factory.line_graph(3, wait_for_announce=True,
opts=[{},
{},
{'disconnect': ['-WIRE_REVOKE_AND_ACK', 'permfail']}])
opts=[{'feerates': (7500, 7500, 7500, 7500)},
# Second commitment_signed is to l3
{'disconnect': ['xWIRE_COMMITMENT_SIGNED*2'],
# We want htlc killed by timeout, not a close due to ping timer.
'dev-no-ping-timer': None},
{'dev-no-ping-timer': None}])
# Start payment
inv = l3.rpc.invoice(msatoshi=10000, label='x', description='desc')['bolt11']
fut = executor.submit(l1.rpc.pay, inv)
ph1 = l3.rpc.invoice(msatoshi="10000sat", label='x1', description='desc2')['payment_hash']
ph2 = l3.rpc.invoice(msatoshi="10000sat", label='x2', description='desc2')['payment_hash']
route = l1.rpc.getroute(l3.info['id'], 1, 1)['route']
# Start a payment
l1.rpc.sendpay(route, ph1)
# l3 sends commitment_signed, then silence.
l2.daemon.wait_for_log('dev_disconnect: xWIRE_COMMITMENT_SIGNED')
# Send another payment, this times out.
l1.rpc.sendpay(route, ph2)
# This can take 30 seconds...
l2.daemon.wait_for_log('Adding HTLC 1 too slow: killing connection',
timeout=TIMEOUT + 30)
l2.daemon.wait_for_log('Failing HTLC 1 due to peer death')
with pytest.raises(RpcError, match=r'WIRE_TEMPORARY_CHANNEL_FAILURE \(reply from remote\)'):
l1.rpc.waitsendpay(ph2, timeout=TIMEOUT)
# l3 closes unilaterally.
wait_for(lambda: only_one(l3.rpc.listpeers(l2.info['id'])['peers'])['connected'] is False)
l3.rpc.close(l2.info['id'], 1)
# l3 goes onchain, without htlc
l3.daemon.wait_for_log('dev_disconnect: -WIRE_REVOKE_AND_ACK')
l3.daemon.wait_for_log('sendrawtransaction')
# Mine it
@ -3659,12 +3682,19 @@ We send an HTLC, and peer unilaterally closes: do we close upstream?
l2.daemon.wait_for_logs(['Their unilateral tx',
r'We want to know if htlc 0 is missing \(later\)'])
# # l1 disconnects now
# l1.rpc.disconnect(l2.info['id'], force=True)
# # Restart now, and reconnect
# l2.restart()
# l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
# After three blocks, onchaind says: definitely missing htlc
bitcoind.generate_block(3)
l2.daemon.wait_for_log('Sending 1 missing htlc messages')
# l2 will tell l1 it has failed the htlc.
# l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
l1.daemon.wait_for_log('peer_in WIRE_UPDATE_FAIL_HTLC')
with pytest.raises(RpcError, match=r'WIRE_PERMANENT_CHANNEL_FAILURE \(reply from remote\)'):
fut.result(TIMEOUT)
l1.rpc.waitsendpay(ph1, timeout=TIMEOUT)