Function test_pay_exemptfee

tests/test_pay.py:3783–3818

Tiny payment, huge fee l1 -> l2 -> l3 Create a tiny invoice for 1 msat, it'll be dominated by the base_fee on the l2->l3 channel. So it'll get rejected on the first attempt if we set the exemptfee way too low. The default fee exemption threshold is 5000msat, so 5001msat is not

(node_factory)

def test_pay_exemptfee(node_factory):
    """Tiny payment, huge fee

    l1 -> l2 -> l3

    Create a tiny invoice for 1 msat, it'll be dominated by the base_fee on
    the l2->l3 channel. So it'll get rejected on the first attempt if we set
    the exemptfee way too low. The default fee exemption threshold is
    5000msat, so 5001msat is not exempted by default and a 5001msat fee on
    l2->l3 should trigger this.

    """
    l1, l2, l3 = node_factory.line_graph(
        3,
        opts=[{}, {'fee-base': 5001, 'fee-per-satoshi': 0}, {}],
        wait_for_announce=True
    )

    err = r'Could not find route without excessive cost'

    with pytest.raises(RpcError, match=err):
        l1.dev_pay(l3.rpc.invoice(1, "lbl1", "desc")['bolt11'], dev_use_shadow=False)

    # If we tell our node that 5001msat is ok this should work
    l1.dev_pay(l3.rpc.invoice(1, "lbl2", "desc")['bolt11'], dev_use_shadow=False, exemptfee=5001)

    # Given the above network this is the smallest amount that passes without
    # the fee-exemption (notice that we let it through on equality).
    threshold = int(5001 / 0.05)

    # This should be just below the fee-exemption and is the first value that is allowed through
    with pytest.raises(RpcError, match=err):
        l1.dev_pay(l3.rpc.invoice(threshold - 1, "lbl3", "desc")['bolt11'], dev_use_shadow=False)

    # While this'll work just fine
    l1.dev_pay(l3.rpc.invoice(int(5001 * 200), "lbl4", "desc")['bolt11'], dev_use_shadow=False)


def test_pay_peer(node_factory, bitcoind):