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feat: subtract dust reserves on the fly from min-capacity-sat

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Michael Schmoock
2019-04-15 22:48:12 +02:00
committed by neil saitug
parent 05b1b3f488
commit 4985693bea
2 changed files with 28 additions and 15 deletions
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30
tests/test_connection.py
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@@ -153,13 +153,13 @@ def test_opening_tiny_channel(node_factory):
min_commit_tx_fees = 5430
min_for_funder = min_commit_tx_fees + dustlimit + 1
l2_min_capacity = 1000 # the old default of 1k sats
l1_min_capacity = 1000 # 1k old default, too small but used at l1 to allow small incoming channels
l2_min_capacity = reserves # just enough to get past capacity filter
l3_min_capacity = min_for_funder # the absolute technical minimum
l4_min_capacity = 10000 # the current default
l5_min_capacity = 20000 # a server with more than default minimum
l4_min_capacity = 10000 # the current default
l5_min_capacity = 20000 # a server with more than default minimum
# Outgoing node must have smallest min value, so inbound side of test channels wont be rejected
l1 = node_factory.get_node(options={'min-capacity-sat': 1000})
l1 = node_factory.get_node(options={'min-capacity-sat': l1_min_capacity})
l2 = node_factory.get_node(options={'min-capacity-sat': l2_min_capacity})
l3 = node_factory.get_node(options={'min-capacity-sat': l3_min_capacity})
l4 = node_factory.get_node(options={'min-capacity-sat': l4_min_capacity})
@@ -170,31 +170,31 @@ def test_opening_tiny_channel(node_factory):
l1.rpc.connect(l4.info['id'], 'localhost', l4.port)
l1.rpc.connect(l5.info['id'], 'localhost', l5.port)
# Open channel with one less than 1000 sats should be rejected at l2
with pytest.raises(RpcError, match=r'channel capacity is .*sat, which is below .*sat'):
l1.fund_channel(l2, l2_min_capacity + reserves - 1)
# Open channel with one less than reserves should be rejected at l2
with pytest.raises(RpcError, match=r'channel_reserve_satoshis .*sat and .*sat too large for funding .*sat'):
l1.fund_channel(l2, l2_min_capacity - 1)
# Open a channel with exactly the minimal amount for the fundee,
# This will raise an exception at l1, as the funder cannot afford fees for initial_commit_tx.
# Note: The old default of 1k sat is below the technical minimum when accounting for dust reserves
# Note: The old default of 1k sat is below the technical minimum when accounting for dust reserves and fees
# This is why this must fail, for this reason the default will be raised to 10k sat.
with pytest.raises(RpcError, match=r'Funder cannot afford fee on initial commitment transaction'):
l1.fund_channel(l2, l2_min_capacity + reserves)
l1.fund_channel(l2, l2_min_capacity)
# Open channel with one less than technical minimum should be rejected at l3
with pytest.raises(RpcError, match=r'channel capacity is .*sat, which is below .*sat'):
l1.fund_channel(l3, l3_min_capacity + reserves - 1)
l1.fund_channel(l3, l3_min_capacity - 1)
# When amount technical minimum matches exactly, own initial_commit_tx fees can now be covered
l1.fund_channel(l3, l3_min_capacity + reserves)
l1.fund_channel(l3, l3_min_capacity)
# Open channel with one less than default 10k sats should be rejected at l4
with pytest.raises(RpcError, match=r'channel capacity is .*, which is below .*msat'):
l1.fund_channel(l4, l4_min_capacity + reserves - 1)
l1.fund_channel(l4, l4_min_capacity - 1)
# This must be possible with enough capacity
l1.fund_channel(l4, l4_min_capacity + reserves)
l1.fund_channel(l4, l4_min_capacity)
# Open channel with less than minimum should be rejected at l5
with pytest.raises(RpcError, match=r'channel capacity is .*, which is below .*msat'):
l1.fund_channel(l5, l5_min_capacity + reserves - 1)
l1.fund_channel(l5, l5_min_capacity - 1)
# bigger channels must not be affected
l1.fund_channel(l5, l5_min_capacity * 10)