pytest: fix weird case in test_closing_different_fees

We set amounts to a list, but then don't use the values except
as a boolean.

Make it a boolean, which should also speed the test up!

Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
This commit is contained in:
Rusty Russell 2021-06-28 13:51:57 +09:30
parent a8d2336ab9
commit c1ad7ac5cb

View file

@ -236,8 +236,8 @@ def test_closing_different_fees(node_factory, bitcoind, executor):
# Default feerate = 15000/11000/7500/1000 # Default feerate = 15000/11000/7500/1000
# It will start at the second number, accepting anything above the first. # It will start at the second number, accepting anything above the first.
feerates = [[20000, 11000, 15000, 7400], [8000, 6000, 1001, 100]] feerates = [[20000, 11000, 15000, 7400], [8000, 6000, 1001, 100]]
amounts = [0, 545999, 546000] balance = [False, True]
num_peers = len(feerates) * len(amounts) num_peers = len(feerates) * len(balance)
addr = l1.rpc.newaddr()['bech32'] addr = l1.rpc.newaddr()['bech32']
bitcoind.rpc.sendtoaddress(addr, 1) bitcoind.rpc.sendtoaddress(addr, 1)
@ -248,10 +248,10 @@ def test_closing_different_fees(node_factory, bitcoind, executor):
# Create them in a batch, for speed! # Create them in a batch, for speed!
peers = [] peers = []
for feerate in feerates: for feerate in feerates:
for amount in amounts: for b in balance:
p = node_factory.get_node(feerates=feerate) p = node_factory.get_node(feerates=feerate)
p.feerate = feerate p.feerate = feerate
p.amount = amount p.balance = balance
l1.rpc.connect(p.info['id'], 'localhost', p.port) l1.rpc.connect(p.info['id'], 'localhost', p.port)
peers.append(p) peers.append(p)
@ -266,7 +266,7 @@ def test_closing_different_fees(node_factory, bitcoind, executor):
l1.daemon.wait_for_logs(['update for channel .* now ACTIVE'] * num_peers l1.daemon.wait_for_logs(['update for channel .* now ACTIVE'] * num_peers
+ ['to CHANNELD_NORMAL'] * num_peers) + ['to CHANNELD_NORMAL'] * num_peers)
for p in peers: for p in peers:
if p.amount != 0: if p.balance:
l1.pay(p, 100000000) l1.pay(p, 100000000)
# Now close all channels (not unilaterally!) # Now close all channels (not unilaterally!)