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Fix the scheduler from dropping UTXOs when there weren't any payments

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Luke Parker 2023-04-13 20:59:36 -04:00
parent 72dd665ebf
commit a5002c50ec
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1 changed files with 14 additions and 0 deletions
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14
processor/src/scheduler.rs
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@ -111,6 +111,8 @@ impl<C: Coin> Scheduler<C> {
// When Substrate emits `Updates` for a coin, all outputs should be added up to the // When Substrate emits `Updates` for a coin, all outputs should be added up to the
// acknowledged block. // acknowledged block.
pub fn add_outputs(&mut self, mut utxos: Vec<C::Output>) -> Vec<Plan<C>> { pub fn add_outputs(&mut self, mut utxos: Vec<C::Output>) -> Vec<Plan<C>> {
log::info!("adding {} outputs", utxos.len());
let mut txs = vec![]; let mut txs = vec![];
for utxo in utxos.drain(..) { for utxo in utxos.drain(..) {
@ -154,9 +156,13 @@ impl<C: Coin> Scheduler<C> {
// If we don't have UTXOs available, don't try to continue // If we don't have UTXOs available, don't try to continue
if self.utxos.is_empty() { if self.utxos.is_empty() {
log::info!("no utxos currently avilable");
return vec![]; return vec![];
} }
// Sort UTXOs so the highest valued ones are first
self.utxos.sort_by(|a, b| a.amount().cmp(&b.amount()).reverse());
// We always want to aggregate our UTXOs into a single UTXO in the name of simplicity // We always want to aggregate our UTXOs into a single UTXO in the name of simplicity
// We may have more UTXOs than will fit into a TX though // We may have more UTXOs than will fit into a TX though
// We use the most valuable UTXOs to handle our current payments, and we return aggregation TXs // We use the most valuable UTXOs to handle our current payments, and we return aggregation TXs
@ -165,6 +171,8 @@ impl<C: Coin> Scheduler<C> {
let utxos = self.utxos.drain(..).collect::<Vec<_>>(); let utxos = self.utxos.drain(..).collect::<Vec<_>>();
let mut utxo_chunks = let mut utxo_chunks =
utxos.chunks(C::MAX_INPUTS).map(|chunk| chunk.to_vec()).collect::<Vec<_>>(); utxos.chunks(C::MAX_INPUTS).map(|chunk| chunk.to_vec()).collect::<Vec<_>>();
// Use the first chunk for any scheduled payments, since it has the most value
let utxos = utxo_chunks.remove(0); let utxos = utxo_chunks.remove(0);
// If the last chunk exists and only has one output, don't try aggregating it // If the last chunk exists and only has one output, don't try aggregating it
@ -179,6 +187,10 @@ impl<C: Coin> Scheduler<C> {
let mut txs = vec![]; let mut txs = vec![];
for chunk in utxo_chunks.drain(..) { for chunk in utxo_chunks.drain(..) {
// TODO: While payments have their TXs' fees deducted from themselves, that doesn't hold here
// We need to charge a fee before reporting incoming UTXOs to Substrate to cover aggregation
// TXs
log::debug!("aggregating a chunk of {} inputs", C::MAX_INPUTS);
txs.push(Plan { key: self.key, inputs: chunk, payments: vec![], change: Some(self.key) }) txs.push(Plan { key: self.key, inputs: chunk, payments: vec![], change: Some(self.key) })
} }
@ -209,6 +221,8 @@ impl<C: Coin> Scheduler<C> {
// for them // for them
if !executing.is_empty() { if !executing.is_empty() {
txs.push(self.execute(utxos, executing)); txs.push(self.execute(utxos, executing));
} else {
self.utxos.extend(utxos);
} }
log::info!( log::info!(