pytest: test persistence of old scids, even if we spliced multiple times.
What the developer wrote
Authored by Rusty Russell
pytest: test persistence of old scids, even if we spliced multiple times.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
What changed, and why it matters
This commit only adds a new test case to the project's test suite. It checks that old channel identifiers (short channel IDs, or SCIDs) remain usable for routing payments even after a channel has been spliced multiple times and a node has restarted. There is no change to production code, no bug fix, and no security-relevant behavior introduced.
No action required. This is a regression test addition and does not alter runtime security posture.
Security signals we found
No strong security signals were identified.
Evidence from the diff
The diff extends tests/test_splicing.py::test_route_by_old_scid with an additional scenario: after a second splice, restarting l2 and reconnecting peers, then verifying that a payment can still be routed using the original pre-splice SCID. The test enables may_reconnect on the nodes and adds a second invoice for the post-restart payment. No implementation code is modified.
Changed components
tests/test_splicing.pyInspect captured patch +27 / −1
diff --git a/tests/test_splicing.py b/tests/test_splicing.py
index 79df7659..aed5061c 100644
--- a/tests/test_splicing.py
+++ b/tests/test_splicing.py
@@ -501,10 +501,11 @@ def test_splice_stuck_htlc(node_factory, bitcoind, executor):
@unittest.skipIf(TEST_NETWORK != 'regtest', 'elementsd doesnt yet support PSBT features we need')
def test_route_by_old_scid(node_factory, bitcoind):
- l1, l2, l3 = node_factory.line_graph(3, wait_for_announce=True, opts={'experimental-splicing': None})
+ l1, l2, l3 = node_factory.line_graph(3, wait_for_announce=True, opts={'experimental-splicing': None, 'may_reconnect': True})
# Get pre-splice route.
inv = l3.rpc.invoice(10000000, 'test_route_by_old_scid', 'test_route_by_old_scid')
+ inv2 = l3.rpc.invoice(10000000, 'test_route_by_old_scid2', 'test_route_by_old_scid2')
route = l1.rpc.getroute(l3.info['id'], 10000000, 1)['route']
# Do a splice
@@ -525,3 +526,28 @@ def test_route_by_old_scid(node_factory, bitcoind):
# Now l1 tries to send using old scid: should work
l1.rpc.sendpay(route, inv['payment_hash'], payment_secret=inv['payment_secret'])
l1.rpc.waitsendpay(inv['payment_hash'])
+
+ # Let's splice again, so the original scid is two behind the times.
+ l3.fundwallet(200000)
+ funds_result = l3.rpc.fundpsbt("109000sat", "slow", 166, excess_as_change=True)
+ chan_id = l3.get_channel_id(l2)
+ result = l3.rpc.splice_init(chan_id, 100000, funds_result['psbt'])
+ result = l3.rpc.splice_update(chan_id, result['psbt'])
+ assert(result['commitments_secured'] is False)
+ result = l3.rpc.splice_update(chan_id, result['psbt'])
+ assert(result['commitments_secured'] is True)
+ result = l3.rpc.signpsbt(result['psbt'])
+ result = l3.rpc.splice_signed(chan_id, result['signed_psbt'])
+
+ wait_for(lambda: only_one(l2.rpc.listpeerchannels(l3.info['id'])['channels'])['state'] == 'CHANNELD_AWAITING_SPLICE')
+ bitcoind.generate_block(6, wait_for_mempool=1)
+ wait_for(lambda: only_one(l2.rpc.listpeerchannels(l3.info['id'])['channels'])['state'] == 'CHANNELD_NORMAL')
+
+ # Now restart l2, make sure it remembers the original!
+ l2.restart()
+ l2.rpc.connect(l1.info['id'], 'localhost', l1.port)
+ l2.rpc.connect(l3.info['id'], 'localhost', l3.port)
+
+ wait_for(lambda: only_one(l1.rpc.listpeers()['peers'])['connected'] is True)
+ l1.rpc.sendpay(route, inv2['payment_hash'], payment_secret=inv2['payment_secret'])
+ l1.rpc.waitsendpay(inv2['payment_hash'])
Why this scored 15/100
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