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pytest: Remove 3 stress-tests

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These tests have proven to be:

 a) very expensive, as they spin up many nodes, and perform long setup
 b) are not testing anything specific, they just fuzz functionality
    that is already tested otherwise
 c) have not helped pinpoint any issues in living memory
 d) are very flaky, making for really bad signal-to-noise, so much
    that devs usually just restart without even looking at the logs
 e) even if we were to look at the logs, we'd be unable to reproduce
    due to the inherent randomness involved in these tests
 f) are really noisy neighbors, causing other tests to flake as well,
    further muddying the water

All in all, these tests are a waste of time, and source of
frustration.

[ Cleaned up python unused imports --RR ]
Changelog-None
This commit is contained in:
Christian Decker
2022-02-25 17:38:54 +01:00
committed by Rusty Russell
parent ec6d91fd54
commit 86b83e473b
2 changed files with 0 additions and 200 deletions
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62
tests/test_closing.py
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@@ -187,68 +187,6 @@ def test_closing_id(node_factory):
wait_for(lambda: not only_one(l2.rpc.listpeers(l1.info['id'])['peers'])['connected'])
@pytest.mark.slow_test
def test_closing_torture(node_factory, executor, bitcoind):
# We set up a fully-connected mesh of N nodes, then try
# closing them all at once.
amount = 10**6
num_nodes = 10 # => 45 channels (36 seconds on my laptop)
if node_factory.valgrind:
num_nodes -= 4 # => 15 (135 seconds)
nodes = node_factory.get_nodes(num_nodes)
# Make sure bitcoind has plenty of utxos
bitcoind.generate_block(num_nodes)
# Give them all plenty of UTXOs, make sure they see them
for i in range(len(nodes)):
for j in range(i + 1, len(nodes)):
addr = nodes[i].rpc.newaddr()['bech32']
bitcoind.rpc.sendtoaddress(addr, (amount + 1000000) / 10**8)
bitcoind.generate_block(1)
sync_blockheight(bitcoind, nodes)
txs = []
for i in range(len(nodes)):
for j in range(i + 1, len(nodes)):
nodes[i].rpc.connect(nodes[j].info['id'], 'localhost', nodes[j].port)
txs.append(nodes[i].rpc.fundchannel(nodes[j].info['id'], amount)['txid'])
# Make sure they're all in, then lock them in.
bitcoind.generate_block(1, wait_for_mempool=txs)
# Wait for them all to be CHANNELD_NORMAL
for n in nodes:
wait_for(lambda: all(p['channels'][0]['state'] == 'CHANNELD_NORMAL' for p in n.rpc.listpeers()['peers']))
# Start closers: can take a long time under valgrind!
futures = []
for i in range(len(nodes)):
for j in range(i + 1, len(nodes)):
futures.append(executor.submit(nodes[i].rpc.close, nodes[j].info['id']))
futures.append(executor.submit(nodes[j].rpc.close, nodes[i].info['id']))
# Wait for close to finish
close_txs = set()
for f in futures:
# If one side completes closing, we'll get an error here 'Peer has no active channel'
try:
close_txs.add(f.result(TIMEOUT)['txid'])
except RpcError as err:
assert err.error['message'] == 'Peer has no active channel'
# Should have one close for each open.
assert len(close_txs) == len(txs)
# Get closes confirmed
bitcoind.generate_block(100, wait_for_mempool=list(close_txs))
# And make sure they hangup.
for n in nodes:
wait_for(lambda: n.rpc.listpeers()['peers'] == [])
@unittest.skipIf(TEST_NETWORK != 'regtest', 'FIXME: broken under elements')
@pytest.mark.slow_test
def test_closing_different_fees(node_factory, bitcoind, executor):