fix: #6558 don't try to repair invalid self-imports. #6573
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Attempts break idempotence.
MarcusGrass
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ytmimi
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Jun 10, 2025
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| // Normalise foo::self -> foo. | ||
| if let UseSegmentKind::Slf(None) = last.kind { | ||
| if !self.path.is_empty() { | ||
| return self; | ||
| } | ||
| } |
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I'd still like to normalize foo::self -> foo. Maybe we can special case this to not normalize self::self -> self to prevent idempotent issues in this really obscure case.
ytmimi
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| UseSegmentKind::Slf(None) if self.path.is_empty() && self.visibility.is_some() => { | ||
| self.path = vec![]; | ||
| return self; | ||
| } |
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I take it this is where we get rid of use self;? I don't think there's any harm in keeping this behavior as long as we handle use self::self in a way that's idempotent.
…ke `self::self`
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A lot fewer changes to behaviour was caused this time around |
|
Will take another look at this later this evening. Thanks for applying the feedback so quickly |
ytmimi
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Jun 11, 2025
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Maybe in a future diff --git a/src/imports.rs b/src/imports.rs
index cfc2f3bd..9f30ac09 100644
--- a/src/imports.rs
+++ b/src/imports.rs
@@ -565,10 +565,18 @@ impl UseTree {
}
// Normalise foo::self -> foo.
+ // Normalise bar::self::self -> bar.
+ // Remove self::self.
if let UseSegmentKind::Slf(None) = last.kind {
- if !self.path.is_empty() {
- return self;
+ while let Some(UseSegmentKind::Slf(_)) = self.path.last().map(|segment| &segment.kind) {
+ self.path.pop();
+ }
+
+ if self.path.is_empty() {
+ self.path = vec![];
}
+
+ return self;
}
// Normalise foo::self as bar -> foo as bar. |
ytmimi
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Fix #6558
Current attempts break idempotence.
As the tests show, paths ending with
selfare now untouched, this PR just removes some code (and adds/modifies some tests).A potential change here would be to keep this part of the code:
In that case
would be removed, but not:
Which woud also fix the idempotence issue, but formatting away
use self;. Although maybe it's best that the compiler inform the user that they're doing something wrong instead of 'sweeping it under the rug' so to say.