It now looks like (for test_hardmpp):
```
# we have computed a set of 1 flows with probability 0.328, fees 0msat and delay 23
# Flow 1: amount=1800000000msat prob=0.328 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0->-103x3x0/1->
# Flow 1: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 1: Failure of 1800000000msat for 103x5x0/0 capacity [0msat,3000000000msat] -> [0msat,1799999999msat]
# we have computed a set of 2 flows with probability 0.115, fees 0msat and delay 23
# Flow 2: amount=500000000msat prob=0.475 fees=0msat delay=12 path=-103x6x0/0(min=max=4294967295msat)->-103x1x0/1->-103x4x0/1->
# Flow 3: amount=1300000000msat prob=0.242 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1799999999msat)->-103x3x0/1->
# Flow 3: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 3: Failure of 1300000000msat for 103x5x0/0 capacity [0msat,1799999999msat] -> [0msat,1299999999msat]
# we have computed a set of 2 flows with probability 0.084, fees 0msat and delay 23
# Flow 4: amount=260000000msat prob=0.467 fees=0msat delay=12 path=-103x6x0/0(500000000msat in 1 htlcs,min=max=4294967295msat)->-103x1x0/1(500000000msat in 1 htlcs)->-103x4x0/1(500000000msat in 1 htlcs)->
# Flow 5: amount=1040000000msat prob=0.179 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1299999999msat)->-103x3x0/1->
# Flow 5: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 5: Failure of 1040000000msat for 103x5x0/0 capacity [0msat,1299999999msat] -> [0msat,1039999999msat]
# we have computed a set of 2 flows with probability 0.052, fees 0msat and delay 23
# Flow 6: amount=120000000msat prob=0.494 fees=0msat delay=12 path=-103x6x0/0(760000000msat in 2 htlcs,min=max=4294967295msat)->-103x1x0/1(760000000msat in 2 htlcs)->-103x4x0/1(760000000msat in 2 htlcs)->
# Flow 7: amount=920000000msat prob=0.105 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1039999999msat)->-103x3x0/1->
# Flow 7: Success
```
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
Unifies the pay_flow resolve functions, and moves remove_htlc_payflow
and commit_htlc_payflow to the top.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
We want to make sure that on every path, we terminate the flow. The simplest
way to do this is encourage the pattern "return pay_flow_xxx(flow)".
Indeed, this caught a few places I missed!
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
The main function here is payment_reconsider:
* Each payment has a list of pay_flow.
* This is populated in try_paying(), calling add_payflows & sendpay_new_flows.
* When we get a notification, we resolve a pay_flow using one of the pay_flow_failedxxx
or pay_flow_succeeded functions.
* They call payment_reconsider() which cleans up finished flows decides what to do:
often calling try_paying again.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
As recommended by your TODO, a bit simpler: we also make the hash function
return a ptr rather than the (now rather large) struct.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
There are a few fields in `struct renepay` which are genuinely
transient, but it makes the code much harder to follow than simply
having a single structure.
More cleanups will follow, but this is the minimal set.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
- adopt "const <type> *"convention
- remove use_shadow option for some pyln tests
- show prob. information of flows into paynotes
- show prob. of success of entire payment flow in paynotes
- minflow: We were not releasing the memory of flow arrays when replacing
them with a new canditate.
- use memleak_scan_obj in memleak_check
- replace u64 with size_t
Signed-off-by: Lagrang3 <eduardo.quintana@pm.me>
