Hiding of member function templates - Which compiler is correct?

Question

I wrote the following code that hiding of member function templates.

#include <iostream>  struct A  {     template<int> void func()     {         std::cout<<"Function tamplate of A"<<std::endl;     } };  struct B : A  {     template<char> void func()     {         std::cout<<"Function tamplate of B"<<std::endl;     }     using A::func; };  int main() {     B().func<0>(); } 

This program working in in Clang compiler. Live demo Clang

But, GCC compiler give an ambiguity error.Live demo GCC

source_file.cpp: In function ‘int main()’: source_file.cpp:22:17: error: call of overloaded ‘func()’ is ambiguous      B().func<0>(); 

So, Which compiler is correct?

Answer 1

Regarding the example in the OP: as pointed out to me by W.F., what matters here is that those are member function templates. You added a using declaration, which specifies ([namespace.udecl]/15):

When a using-declaration brings names from a base class into a derived class scope, member functions and member function templates in the derived class override and/or hide member functions and member function templates with the same name, parameter-type-list, cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting).

Note how template parameters aren't accounted. And it is Clang that treated the code correctly by hiding the int version.

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On the other hand if one examines the example tobi303 suggested under your post, GCC is kinda in the right. This simply isn't specified to be resolved somehow.

For one, there's [temp.fct.spec]/3:

Trailing template arguments that can be deduced or obtained from default template-arguments may be omitted from the list of explicit template-arguments. [...] In contexts where deduction is done and fails, or in contexts where deduction is not done, if a template argument list is specified and it, along with any default template arguments, identifies a single function template specialization, then the template-id is an lvalue for the function template specialization.

The text in bold indicates that your program is well-formed only if the template argument we give nominates a single specialization. And ostensibly, it doesn't, since according to [temp.arg.nontype]/1:

A template-argument for a non-type, non-template template-parameter shall be one of:

• for a non-type template-parameter of integral or enumeration type, a converted constant expression of the type of the template-parameter;

And 0 fits both overloads as a converted constant expression. On account of there not being any ICS ranking for template arguments, this is ambiguous.