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pytest: change multihtlc topology for simpler testing. 

We were seeing flakes due to channel closures on intermidiary nodes if
the blocks came too fast.

If we use a star topology it's actually what we want and it's much simpler to
figure out what's going on.

Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
This commit is contained in:
Rusty Russell 2022-07-30 14:00:40 +09:30 committed by neil saitug
parent b5ee5e7fb1
commit 93303ffdad
1 changed files with 100 additions and 45 deletions
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145
tests/test_closing.py
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@ -2757,80 +2757,135 @@ def test_permfail_new_commit(node_factory, bitcoind, executor):
def setup_multihtlc_test(node_factory, bitcoind):
# l1 -> l2 -> l3 -> l4 -> l5 -> l6 -> l7
# l1 --\ /-> l6
# v /
# l2 -> l4 -> l5 -> l7
# ^
# l3 --/
# l1 and l7 ignore and HTLCs they're sent.
# For each direction, we create these HTLCs with same payment_hash:
# 1 failed (CLTV1)
# 1 failed (CLTV2)
# 2 live (CLTV2)
# 1 live (CLTV3)
nodes = node_factory.line_graph(7, wait_for_announce=True,
opts={'dev-no-reconnect': None,
'may_reconnect': True})
l1, l2, l3, l4, l5, l6, l7 = node_factory.get_nodes(7,
opts={'dev-no-reconnect': None,
'may_reconnect': True})
# Balance by pushing half the funds.
b11 = nodes[-1].rpc.invoice(10**9 // 2, '1', 'balancer')['bolt11']
nodes[0].rpc.pay(b11)
l4.fundwallet(10**6 * 4 + 100000)
l5.fundwallet(10**6 * 2 + 100000)
nodes[0].rpc.dev_ignore_htlcs(id=nodes[1].info['id'], ignore=True)
nodes[-1].rpc.dev_ignore_htlcs(id=nodes[-2].info['id'], ignore=True)
# They need to be connected.
for n in l1, l2, l3, l5:
l4.rpc.connect(n.info['id'], 'localhost', n.port)
for n in l6, l7:
l5.rpc.connect(n.info['id'], 'localhost', n.port)
# Efficient way to establish the channels.
l4.rpc.multifundchannel([{'id': l1.info['id'], 'amount': 10**6},
{'id': l2.info['id'], 'amount': 10**6},
{'id': l3.info['id'], 'amount': 10**6},
{'id': l5.info['id'], 'amount': 10**6}])
l5.rpc.multifundchannel([{'id': l6.info['id'], 'amount': 10**6},
{'id': l7.info['id'], 'amount': 10**6}])
# Make sure they're all in normal state.
bitcoind.generate_block(1)
wait_for(lambda: all([only_one(p['channels'])['state'] == 'CHANNELD_NORMAL'
for p in l4.rpc.listpeers()['peers']]))
wait_for(lambda: all([only_one(p['channels'])['state'] == 'CHANNELD_NORMAL'
for p in l5.rpc.listpeers()['peers']]))
# Balance them
for n in l1, l2, l3, l5:
l4.pay(n, 10**9 // 2)
for n in l6, l7:
l5.pay(n, 10**9 // 2)
def route_to_l7(src):
"""Route from l1, l2 or l3 to l7"""
return [{'id': l4.info['id'], 'channel': src.get_channel_scid(l4), 'amount_msat': "100002002msat", 'delay': 21},
{'id': l5.info['id'], 'channel': l4.get_channel_scid(l5), 'amount_msat': "100001001msat", 'delay': 15},
{'id': l7.info['id'], 'channel': l5.get_channel_scid(l7), 'amount_msat': "100000000msat", 'delay': 9}]
def route_to_l1(src):
"""Route from l6 or l7 to l1"""
return [{'id': l5.info['id'], 'channel': src.get_channel_scid(l5), 'amount_msat': "100002002msat", 'delay': 21},
{'id': l4.info['id'], 'channel': l5.get_channel_scid(l4), 'amount_msat': "100001001msat", 'delay': 15},
{'id': l1.info['id'], 'channel': l4.get_channel_scid(l1), 'amount_msat': "100000000msat", 'delay': 9}]
# Freeze the HTLCs in place.
l1.rpc.dev_ignore_htlcs(id=l4.info['id'], ignore=True)
l7.rpc.dev_ignore_htlcs(id=l5.info['id'], ignore=True)
preimage = "0" * 64
inv = nodes[0].rpc.invoice(amount_msat=10**8, label='x', description='desc',
preimage=preimage)
inv = l1.rpc.invoice(amount_msat=10**8, label='x', description='desc',
preimage=preimage)
h = inv['payment_hash']
nodes[-1].rpc.invoice(amount_msat=10**8, label='x', description='desc',
preimage=preimage)['payment_hash']
l7.rpc.invoice(amount_msat=10**8, label='x', description='desc',
preimage=preimage)['payment_hash']
# First, the failed attempts (paying wrong node). CLTV1
r = nodes[0].rpc.getroute(nodes[-2].info['id'], 10**8, 1)["route"]
nodes[0].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l7(l1)
r[2]['id'] = l6.info['id']
r[2]['channel'] = l5.get_channel_scid(l6)
l1.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
with pytest.raises(RpcError, match=r'INCORRECT_OR_UNKNOWN_PAYMENT_DETAILS'):
nodes[0].rpc.waitsendpay(h)
l1.rpc.waitsendpay(h)
r = nodes[-1].rpc.getroute(nodes[1].info['id'], 10**8, 1)["route"]
nodes[-1].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l1(l7)
r[2]['id'] = l2.info['id']
r[2]['channel'] = l4.get_channel_scid(l2)
l7.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
with pytest.raises(RpcError, match=r'INCORRECT_OR_UNKNOWN_PAYMENT_DETAILS'):
nodes[-1].rpc.waitsendpay(h)
l7.rpc.waitsendpay(h)
# Now increment CLTV -> CLTV2
bitcoind.generate_block(1)
sync_blockheight(bitcoind, nodes)
sync_blockheight(bitcoind, (l1, l2, l3, l4, l5, l6, l7))
# Now, the live attempts with CLTV2 (blackholed by end nodes)
r = nodes[0].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
nodes[0].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = nodes[-1].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
nodes[-1].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l7(l1)
l1.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l1(l7)
l7.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
# We send second HTLC from different node, since they refuse to send
# multiple with same hash.
r = nodes[1].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
nodes[1].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = nodes[-2].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
nodes[-2].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l7(l2)
l2.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l1(l6)
l6.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
# Now increment CLTV -> CLTV3.
bitcoind.generate_block(1)
sync_blockheight(bitcoind, nodes)
sync_blockheight(bitcoind, (l1, l2, l3, l4, l5, l6, l7))
r = nodes[2].rpc.getroute(nodes[-1].info['id'], 10**8, 1)["route"]
nodes[2].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = nodes[-3].rpc.getroute(nodes[0].info['id'], 10**8, 1)["route"]
nodes[-3].rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
r = route_to_l7(l3)
l3.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
# Final HTLC is actually from l5 itself...
r = route_to_l1(l7)[1:]
l5.rpc.sendpay(r, h, payment_secret=inv['payment_secret'])
# Make sure HTLCs have reached the end.
nodes[0].daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 3)
nodes[-1].daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 3)
# Make sure HTLCs have reached the ends (including balance payment!)
l7.daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 4)
l1.daemon.wait_for_logs(['peer_in WIRE_UPDATE_ADD_HTLC'] * 4)
return h, nodes
# We have 6 HTLCs trapped in l4-l5 channel.
assert len(only_one(only_one(l4.rpc.listpeers(l5.info['id'])['peers'])['channels'])['htlcs']) == 6
# We are all connected.
for n in l1, l2, l3, l4, l5, l6, l7:
assert all([p['connected'] for p in n.rpc.listpeers()['peers']])
return h, l1, l2, l3, l4, l5, l6, l7
@pytest.mark.developer("needs DEVELOPER=1 for dev_ignore_htlcs")
@pytest.mark.slow_test
def test_onchain_multihtlc_our_unilateral(node_factory, bitcoind):
"""Node pushes a channel onchain with multiple HTLCs with same payment_hash """
h, (l1, l2, l3, l4, l5, l6, l7) = setup_multihtlc_test(node_factory, bitcoind)
h, l1, l2, l3, l4, l5, l6, l7 = setup_multihtlc_test(node_factory, bitcoind)
# Now l4 goes onchain with l4-l5 channel.
l4.rpc.dev_fail(l5.info['id'])
@ -2850,8 +2905,8 @@ def test_onchain_multihtlc_our_unilateral(node_factory, bitcoind):
l7.restart()
# We disabled auto-reconnect so we'd detect breakage, so manually reconnect.
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
l7.rpc.connect(l6.info['id'], 'localhost', l6.port)
l1.rpc.connect(l4.info['id'], 'localhost', l4.port)
l7.rpc.connect(l5.info['id'], 'localhost', l5.port)
# Wait for HTLCs to stabilize.
l1.daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
@ -2879,14 +2934,14 @@ def test_onchain_multihtlc_our_unilateral(node_factory, bitcoind):
# No other channels should have failed.
for n in l1, l2, l3, l6, l7:
assert all([p['connected'] for p in n.rpc.listpeers()['peers']])
assert only_one(n.rpc.listpeers()['peers'])['connected']
@pytest.mark.developer("needs DEVELOPER=1 for dev_ignore_htlcs")
@pytest.mark.slow_test
def test_onchain_multihtlc_their_unilateral(node_factory, bitcoind):
"""Node pushes a channel onchain with multiple HTLCs with same payment_hash """
h, (l1, l2, l3, l4, l5, l6, l7) = setup_multihtlc_test(node_factory, bitcoind)
h, l1, l2, l3, l4, l5, l6, l7 = setup_multihtlc_test(node_factory, bitcoind)
# Now l5 goes onchain with l4-l5 channel.
l5.rpc.dev_fail(l4.info['id'])
@ -2906,8 +2961,8 @@ def test_onchain_multihtlc_their_unilateral(node_factory, bitcoind):
l7.restart()
# We disabled auto-reconnect so we'd detect breakage, so manually reconnect.
l1.rpc.connect(l2.info['id'], 'localhost', l2.port)
l7.rpc.connect(l6.info['id'], 'localhost', l6.port)
l1.rpc.connect(l4.info['id'], 'localhost', l4.port)
l7.rpc.connect(l5.info['id'], 'localhost', l5.port)
# Wait for HTLCs to stabilize.
l1.daemon.wait_for_logs(['peer_out WIRE_UPDATE_FAIL_HTLC'] * 3)
@ -2936,7 +2991,7 @@ def test_onchain_multihtlc_their_unilateral(node_factory, bitcoind):
# No other channels should have failed.
for n in l1, l2, l3, l6, l7:
assert all([p['connected'] for p in n.rpc.listpeers()['peers']])
assert only_one(n.rpc.listpeers()['peers'])['connected']
@pytest.mark.developer("needs DEVELOPER=1")