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Revert back to Vec to hold utxos in builder

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Due to brain malfunction I made utxos into a BTree. This made a test
pass but is incorrect. The test itself was incorrect as per comment in

https://github.com/bitcoindevkit/bdk/pull/258#issuecomment-758370380

So I (1) reverted utxos back to a Vec, (2) fixed the test and expanded
the comment in the test.
This commit is contained in:
LLFourn 2021-01-22 14:04:06 +11:00
parent 890d6191a1
commit 10fcba9439
No known key found for this signature in database
GPG Key ID: A27093B54DA11F65
2 changed files with 14 additions and 19 deletions
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17
src/wallet/mod.rs
View File

@ -507,7 +507,7 @@ where
let (required_utxos, optional_utxos) = self.preselect_utxos(
params.change_policy,
&params.unspendable,
params.utxos.values().cloned().collect(),
params.utxos.clone(),
params.drain_wallet,
params.manually_selected_only,
params.bumping_fee.is_some(), // we mandate confirmed transactions if we're bumping the fee
@ -672,7 +672,7 @@ where
let original_txin = tx.input.drain(..).collect::<Vec<_>>();
let original_utxos = original_txin
.iter()
.map(|txin| -> Result<(OutPoint, (UTXO, usize)), Error> {
.map(|txin| -> Result<_, Error> {
let txout = self
.database
.borrow()
@ -706,9 +706,9 @@ where
keychain,
};
Ok((utxo.outpoint, (utxo, weight)))
Ok((utxo, weight))
})
.collect::<Result<BTreeMap<OutPoint, (UTXO, usize)>, _>>()?;
.collect::<Result<Vec<_>, _>>()?;
if tx.output.len() > 1 {
let mut change_index = None;
@ -2653,9 +2653,10 @@ mod test {
fn test_bump_fee_remove_output_manually_selected_only() {
let (wallet, descriptors, _) = get_funded_wallet(get_test_wpkh());
// receive an extra tx so that our wallet has two utxos. then we manually pick only one of
// them, and make sure that `bump_fee` doesn't try to add more. eventually, it should fail
// because the fee rate is too high and the single utxo isn't enough to create a non-dust
// output
// them, and make sure that `bump_fee` doesn't try to add more. This fails because we've
// told the wallet it's not allowed to add more inputs AND it can't reduce the value of the
// existing output. In other words, bump_fee + manually_selected_only is always an error
// unless you've also set "maintain_single_recipient".
let incoming_txid = crate::populate_test_db!(
wallet.database.borrow_mut(),
testutils! (@tx ( (@external descriptors, 0) => 25_000 ) (@confirmations 1)),
@ -2694,8 +2695,6 @@ mod test {
let mut builder = wallet.build_fee_bump(txid).unwrap();
builder
.add_utxo(outpoint)
.unwrap()
.manually_selected_only()
.fee_rate(FeeRate::from_sat_per_vb(255.0));
builder.finish().unwrap();

16
src/wallet/tx_builder.rs
View File

@ -147,7 +147,7 @@ pub(crate) struct TxParams {
pub(crate) fee_policy: Option<FeePolicy>,
pub(crate) internal_policy_path: Option<BTreeMap<String, Vec<usize>>>,
pub(crate) external_policy_path: Option<BTreeMap<String, Vec<usize>>>,
pub(crate) utxos: BTreeMap<OutPoint, (UTXO, usize)>,
pub(crate) utxos: Vec<(UTXO, usize)>,
pub(crate) unspendable: HashSet<OutPoint>,
pub(crate) manually_selected_only: bool,
pub(crate) sighash: Option<SigHashType>,
@ -273,15 +273,11 @@ impl<'a, B, D: BatchDatabase, Cs: CoinSelectionAlgorithm<D>, Ctx: TxBuilderConte
/// These have priority over the "unspendable" utxos, meaning that if a utxo is present both in
/// the "utxos" and the "unspendable" list, it will be spent.
pub fn add_utxo(&mut self, outpoint: OutPoint) -> Result<&mut Self, Error> {
if self.params.utxos.get(&outpoint).is_none() {
let deriv_ctx = crate::wallet::descriptor_to_pk_ctx(self.wallet.secp_ctx());
let utxo = self.wallet.get_utxo(outpoint)?.ok_or(Error::UnknownUTXO)?;
let descriptor = self.wallet.get_descriptor_for_keychain(utxo.keychain);
let satisfaction_weight = descriptor.max_satisfaction_weight(deriv_ctx).unwrap();
self.params
.utxos
.insert(outpoint, (utxo, satisfaction_weight));
}
let deriv_ctx = crate::wallet::descriptor_to_pk_ctx(self.wallet.secp_ctx());
let utxo = self.wallet.get_utxo(outpoint)?.ok_or(Error::UnknownUTXO)?;
let descriptor = self.wallet.get_descriptor_for_keychain(utxo.keychain);
let satisfaction_weight = descriptor.max_satisfaction_weight(deriv_ctx).unwrap();
self.params.utxos.push((utxo, satisfaction_weight));
Ok(self)
}