[PATCH] Fix return type deduction for member templates

[Richard Smith]

Hi Faisal,

On Thu, Jun 13, 2013 at 11:30 AM, Faisal Vali <faisalv at gmail.com> wrote:
> While implementing return type deduction for generic lambdas, I stumbled upon
> a bug which led to clang mishandling the following code:
>
> struct Lambda {
>   template<class T> auto operator()(T t) {
>     return 5;
>   }
> template<class T> auto foo(T t) { return t; }
> };
> void test() {
>     Lambda L;
>     int i = L(3);
> }
>
>
> The issue (as i understand it) was that in some contexts only the
> function template (and not the specialization) would get passed to
> DiagnoseUseOfDecl (which deduces the return type of the function).  So
> the specialization would flow through to Codegen with auto as its
> return type.
>
> So, with that in mind, this patch implements a trivial fix.
>
> What do you think Richard? Is there a better way to address this bug?

I was aware of this issue, but got snowed under with other things.
This bug is not limited to 'auto' type deduction; the other code in
DiagnoseUseOfDecl suffers the same way. For instance:

struct Lambda {
  template<class T> static __attribute__((unused)) int foo(T) {
    return 5;
  }
};
int bar() {
    Lambda L;
    return L.foo(3);
}

... does not warn that the function is marked 'unused' but is used,
but if you call it as Lambda::foo(3), it does warn.

I think the right fix here is to pass both the declaration found by
name lookup and the declaration that is actually used into
DiagnoseUseOfDecl.

> Are there other places this could happen (ie. template conversion operator?)

I think there were, but I don't remember. Another thing to try is
hiding the function behind a 'using' declaration.