Highlight struct field access as a field - #1378
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The right-hand side of a member access like `p.x` is not a resolved item nor a pattern, so it was left unhighlighted (or painted as the macro inside an inline macro). Resolve it through the semantic member access expression and highlight it as a field. Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>
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| /// Resolves the right-hand side of a struct member access expression (e.g. the `x` in `p.x`) | ||
| /// to a [`SemanticTokenKind::Field`]. | ||
| fn from_member_access( |
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This is essentially a copy of what we already have in finder.rs:409 (try_member function). Please try to reuse that, i think it's applicable here.
| let expr_id = db.lookup_expr_by_ptr(function_id, binary_expr.stable_ptr(db).into()).ok()?; | ||
| let member_access = match db.expr_semantic(function_id, expr_id) { | ||
| Expr::MemberAccess(member_access) => member_access, | ||
| Expr::Snapshot(snapshot) => match db.expr_semantic(function_id, snapshot.inner) { |
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we should probably have tests that cover snapshots behavior if we support those paths.
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The right-hand side of a member access like
p.xis not a resolved item nor apattern, so it was left unhighlighted (or painted as the macro inside an inline
macro). Resolve it through the semantic member access expression and highlight it
as a field.