Question

#include <iostream>

template <class T> 
void foo(T) {
    std::cout << "foo(T)" << std::endl;
}

template <class T> 
void foo(T*) { //#3
    std::cout << "foo(T*)" << std::endl;
}

#define TEST

#ifdef TEST
template <> 
void foo(int*) { //#1
    std::cout << "foo(int*)" << std::endl;
}
#else
template <>
void foo<int*>(int*) { //#2
    std::cout << "foo<int*>(int*)" << std::endl;
}
#endif

int main(int argc, char **argv) {
    int* p = 0;
    foo(p);
    return 0;
}

What is the difference between #1 and #2 above? If I define TEST, then #1 works, but if I comment it out, #3 works. Which is the right way to write function template specialization here?

Answer 1

#1 declares a function template specialization of #3 and automatically deduces the template parameters. #2 is a specialization of the first template you defined (the one without number, let’s call it #0) for T=int*. It can’t be a specialization of #3 because replacing T with the specified int* there would lead to a int** parameter.

When you call foo, overload resolution now first picks the best fitting base template, then checks that template for any existing specializations. With TEST defined, there are two base templates (#0 and #3) and #3 is a better match and gets selected. Then the compiler checks for specializations of that template, and #1 is a better fit and is being called.

Without TEST defined, there are still two base templates (#0 and #3) and #3 is a better match and gets selected. Then the compiler checks for specializations of that template, but since #2 specializes #0 and not #3, it is not considered and #3 ends of being called.

This is the classical example of Why not Specialize Function Templates. The problems are explained in more detail there.

The simple solution is to not specialize function templates at all, but simply add new overloads for the special types:

// no template, just a normal function
void foo(int*) {
    std::cout << "foo(int*)" << std::endl;
}