Question

#include <iostream>

using namespace std;

     template <typename T> void compare(const T&, const T&){
        cout<<"T"<<endl;
     }
     void compare(const char*, const char*){

        cout<<"const char*"<<endl;
     }

int main()
{

     char a[]="123";

     char b[]="123";

     char *p1 = a, *p2 = b;

     compare(p1,p2);
    return 0;
}

the result is: T

but why? after instantiation the template function may be so:

compare(const char*&, const char*&)

is the same as ordinary function. and the ordinary function should be called!

Answer 1

This is why I think writing T const& is superior to writing const T&.

The template function is instantiated with the signature void(char* const&, char* const&) which is a better match than void(char const*, char const*), as it requires no conversions of the pointers from char* to char const*.