Function test_funding_fail

tests/test_connection.py:1048–1088

Add some funds, fund a channel without enough funds

(node_factory, bitcoind)

@pytest.mark.openchannel('v1')
@pytest.mark.openchannel('v2')
def test_funding_fail(node_factory, bitcoind):
    """Add some funds, fund a channel without enough funds"""
    # Previous runs with same bitcoind can leave funds!
    max_locktime = 5 * 6 * 24
    l1 = node_factory.get_node(random_hsm=True, options={'max-locktime-blocks': max_locktime})
    l2 = node_factory.get_node(options={'watchtime-blocks': max_locktime + 1})
    l1.rpc.connect(l2.info['id'], 'localhost', l2.port)

    funds = 1000000

    addr = l1.rpc.newaddr()['bech32']
    l1.bitcoin.rpc.sendtoaddress(addr, funds / 10**8)
    bitcoind.generate_block(1)

    # Wait for it to arrive.
    wait_for(lambda: len(l1.rpc.listfunds()['outputs']) > 0)

    # Fail because l1 dislikes l2's huge locktime.
    with pytest.raises(RpcError, match=r'to_self_delay \d+ larger than \d+'):
        l1.rpc.fundchannel(l2.info['id'], int(funds / 10))

    # channels disconnect on failure
    wait_for(lambda: len(l1.rpc.listpeers()['peers']) == 0)
    wait_for(lambda: len(l2.rpc.listpeers()['peers']) == 0)

    # Restart l2 without ridiculous locktime.
    del l2.daemon.opts['watchtime-blocks']
    l2.restart()
    l1.rpc.connect(l2.info['id'], 'localhost', l2.port)

    # We don't have enough left to cover fees if we try to spend it all.
    with pytest.raises(RpcError, match=r'not afford'):
        l1.rpc.fundchannel(l2.info['id'], funds)

    # Should still be connected (we didn't contact the peer)
    assert only_one(l1.rpc.listpeers()['peers'])['connected']
    l2.daemon.wait_for_log('Handed peer, entering loop')
    assert only_one(l2.rpc.listpeers()['peers'])['connected']

    # This works.
    l1.rpc.fundchannel(l2.info['id'], int(funds / 10))


@pytest.mark.openchannel('v1')