Template specialization for a compile-time constant integer

Question

Is it possible to specialize a function (or a class at least) to select between constant (compile-time!) integer and all other arguments? And if so, it would be nice to specialize (enable_if) for specific constant values only.

In the example below this would mean output "var", "const", and "var", not three "var"s.

#include <type_traits>
#include <iostream>
using namespace std;

struct test
{
    template <typename T>
    test& operator=(const T& var) { cout << "var" << endl; return *this; }
    template <int n> // enable_if< n == 0 >
    test& operator=(int x) { cout << "const" << endl; return *this; }
};

int main()
{
    test x;
    x = "x";
    x = 1;
    int y = 55;
    x = y;
    return 0;
}

UPDATE: edited the code to emphasize that it has to be compile-time constant.

The only way you could call that second overload is with x.operator=<42>(1);. — Xeo
No I can't do this (because I adapt my class to foreign code). — queen3

Mateusz Pusz Mateusz Pusz · Accepted Answer · 2012-11-16T19:42:11

To get var, const, var in your example you can do that.

struct test {
  template <typename T>
  test& operator=(const T& var) { cout << "var" << endl; return *this; }
  test& operator=(int &&x) { cout << "const" << endl; return *this; }
};

It will work for all temporaries, but it will fail for:

const int yy = 55;
x = yy;

To make it work for such a case too you will need to add:

test& operator=(int &x)       { cout << "var" << endl; return *this; }
test& operator=(const int &x) { cout << "const" << endl; return *this; }

Template specialization for a compile-time constant integer

1 Answers