ilya-biryukov added a comment.
Sorry for delays with this patch, I somehow missed the email notification about
it.
> Expounding on my concerns a bit -- I'm worried about all the other places
> calling isUndeducedType() and whether they're also at risk and thus a better
> fix is to change isUndeducedType() to pay attention to the language mode.
I thought semantics of `isUndeducedType()` is actually what's most callers
(e.g. diagnostics) want. That is "the type is undeduced **and** not dependent".
But the name can definitely bring some confusion, so there might be other
places that misinterpreted it.
@rsmith, it would be great if you could take another look when you have time.
================
Comment at: lib/Sema/SemaDecl.cpp:12609-12610
return false;
+ // We can't simply call Type::isUndeducedType here, because it returns
+ // false on C++14's auto return type without trailing return type.
+ DeducedType *DT = FD->getReturnType()->getContainedDeducedType();
----------------
rsmith wrote:
> I don't think this comment captures the problem. Rather, I think the problem
> is that in a template we deduce the `auto` to a dependent type, so it's not
> considered "undeduced" at the point when we ask this question.
Thanks! That definitely explains the problem in a proper manner.
================
Comment at: lib/Sema/SemaDecl.cpp:12612-12613
+ DeducedType *DT = FD->getReturnType()->getContainedDeducedType();
+ if (DT && DT->getDeducedType().isNull())
+ return false;
+ }
----------------
rsmith wrote:
> Is there any need to check whether the type has been deduced here? It seems
> simpler (and equivalent) to turn off skipping for functions whose return type
> involves a deduced type regardless of whether deduction has been performed
> already (which it hasn't).
Thanks! I misinterpreted the deduced type's semantics. I thought it also
appears in the following case:
`auto foo() -> int {}`
But it doesn't.
Repository:
rC Clang
https://reviews.llvm.org/D44480
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