模板类成员函数的专业化

Question

there:

What is the result of the following codes?

foo.h

#include <iostream>

template<class _Tp>
struct Foo
{
   void print() { std::cout << "foo\n"; }
};

foo.cxx

#include "foo.h"

template<>
void Foo<int>::print()
{
   std::cout << "foo<int>\n";
}

main.cxx

#include "foo.h"

int main()
{
   Foo<double>().print();
   Foo<int>().print();
   return 0;
 }

The results are different:

1. when complied by MSVC,

    foo foo 

2. when compiled by g++,

    foo foo<int> 

I would like to get the second result regardless of compilers. What should I do additionally to achieve? If possible, would you give me an explanation about underlying standards or mechanism. Thank you!

Answer 1

Your program has undefined behavior.

There are two implementations of Foo<int>::print() -- an inline definition that is obtained from the class template definition and the non-inline definition in foo.cxx. A compiler is free to choose either.

A compiler is not required to diagnose that as a problem. Many of them (apparently that includes g++ and MSVC) choose that route for definitions of class templates and their member functions.

To make sure that both compilers choose the implementation in foo.cxx, declare the function in foo.h.

#include <iostream>

template<class _Tp>
struct Foo
{
   void print() { std::cout << "foo\n"; }
};

template<> void Foo<int>::print();

Answer 2

It is interest. I know that templates are interpreted at compile time, not link time. So, in my opinion, specialization template should implements at each cpp files. What is your g++ version?

Answer 3

If the foo.h looks like following, both compilers produce the same results. But I don't know why.

#include <iostream>


template<class _Tp>
struct Foo
{
   void print();
};

template<class _Tp>
void Foo<_Tp>::print()
{
   std::cout << "foo\n";
}

template<> void Foo<int>::print();