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trees: make simplify_conflict() always return a Conflict

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The interface is much cleaner this way. The function no longer makes
any writes to the store.
This commit is contained in:
Martin von Zweigbergk 2021-10-20 22:00:52 -07:00
parent ffa727eb51
commit f1e2124a64
1 changed files with 14 additions and 16 deletions
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30
lib/src/tree.rs
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@ -634,7 +634,17 @@ fn merge_tree_value(
value: side2.clone(),
});
}
simplify_conflict(store, &conflict)?
let conflict = simplify_conflict(store, &conflict)?;
if conflict.adds.is_empty() {
// If there are no values to add, then the path doesn't exist
None
} else if conflict.removes.is_empty() && conflict.adds.len() == 1 {
// A single add means that the current state is that state.
Some(conflict.adds[0].value.clone())
} else {
let conflict_id = store.write_conflict(&conflict)?;
Some(TreeValue::Conflict(conflict_id))
}
}
}
}
@ -659,7 +669,7 @@ fn conflict_part_to_conflict(store: &Store, part: &ConflictPart) -> Result<Confl
fn simplify_conflict(
store: &Store,
conflict: &Conflict,
) -> Result<Option<TreeValue>, BackendError> {
) -> Result<Conflict, BackendError> {
// Important cases to simplify:
//
// D
@ -738,20 +748,8 @@ fn simplify_conflict(
// {+A+A}, that would become just {+A}. Similarly {+B-A+B} would be just
// {+B-A}.
if new_adds.is_empty() {
// If there are no values to add, then the path doesn't exist (so return None to
// indicate that).
return Ok(None);
}
if new_removes.is_empty() && new_adds.len() == 1 {
// A single add means that the current state is that state.
return Ok(Some(new_adds[0].value.clone()));
}
let conflict_id = store.write_conflict(&Conflict {
Ok(Conflict {
adds: new_adds,
removes: new_removes,
})?;
Ok(Some(TreeValue::Conflict(conflict_id)))
})
}