================
@@ -936,19 +940,21 @@ Expr *buildIsDeducibleConstraint(Sema &SemaRef,
Context.DeclarationNames.getCXXDeductionGuideName(AliasTemplate));
};
+ TemplateDecl *TD = DeducingTemplate ? DeducingTemplate : AliasTemplate;
+
SmallVector<TypeSourceInfo *> IsDeducibleTypeTraitArgs = {
Context.getTrivialTypeSourceInfo(
Context.getDeducedTemplateSpecializationType(
- TemplateName(AliasTemplate), /*DeducedType=*/QualType(),
+ TemplateName(TD), /*DeducedType=*/QualType(),
----------------
hokein wrote:
```cpp
template <typename T> struct B {
B(T);
};
template <typename T> struct C : public B<T> {
using B<T>::B;
};
C c(1);
// implicit code
// template <typename> class CC;
//
// template <typename T>
// class CC<B<T>> {
// public:
// typedef C<T> type;
// };
```
If I read the code correctly, we will build the `__is_deducible(C, typename
CC<B<T>>::type)` for the deduction guide of `C` (`auto (T) -> typename
CC<B<T>>::type`). I'm not sure this is correct.
Intuitively, this constraint is not necessary, if `CC<B<T>>::type` is valid, it
must be the type of `C<T>`, then `is_deducible(C, C<T>)` is always true.
Unfortunately, the standard doesn't specify the transformation of the
associated constraint (oversight?). If we only replace the return type, then
the constraint will be `is_deducible(B, typename CC<B<T>>::type)` which is
always false.
Perhaps we should always drop the `is_deducible` constraint for this case.
https://github.com/llvm/llvm-project/pull/98788
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