Question

Following code generate no compilation/linker error/warning:

// A.h
#include<iostream>
struct A
{
  template<typename T>
  static void foo (T t)
  {
    std::cout << "A::foo(T)\n";
  }
};
void other ();

// main.cpp
#include"A.h"
int main ()
{
  A::foo(4.7);
  other();
}

// other.cpp
#include"A.h"
template<>
void A::foo (double d)
{
  cout << "A::foo(double)\n";
}

int other ()
{
  A::foo(4.7);
}

The output surprisingly is:

A::foo(T)
A::foo(double)

Why compiler is not able to pick up the correct A::foo(double) in case of main.cpp ?

Agree that, there is no issue as expected, if there is a declaration in A.h like below:

template<> void A::foo (double);

But that’s not the concern, because at link time, compiler has the specialized version.

Also, is having 2 different version of the same function an Undefined Behavior ?

Answer 1

All explicit specialization declarations must be visible at the time of the template instantiation. Since your explicit specialization declaration for A::foo<double> is visible in one translation unit but not the other, the program is ill-formed.

(In practice, the compiler will instantiate the primary template in main.cpp and the explicitly-specialized one in other.cpp. That would still an ODR violation anyway.)