Template function specialization vs. overloading

Question

From some slides about template specialization:

#include <iostream>

using namespace std;

template<class X> 
X& min(X& a, X& b)
{
    return a > b ? b : a;
}

int& min(int& a, int & b)
{
    // rewrite of the function in the case of int:
    cout << "int explicit function\n";
    return a > b ? b : a;
}

/* 
new syntax – the more appropriate way:
template<>
int& min<int>(int& a, int& b)
{
    cout << "int explicit function\n";
    return a > b ? b : a;
}
*/

Why is the second way more "appropriate"?

Relevant reading: [Why not specialize function templates?](http://www.gotw.ca/publications/mill17.htm) — juanchopanza, Jun 27 '13 at 13:27
Remark: when declaring a template function specialization, you can use template argument deduction. — curiousguy, Jun 28 '13 at 00:38

score 1 · Accepted Answer · answered Jun 27 '13 at 13:28

1

The overload works fine for most of the contexts, and AFAIK is the suggested baseline approach. (see GOTW suggested by juanchopanza )

The difference hits if someone explicitly asks for the template, calling min<int>(x, y). In that case overloads are ignored and only the template (base or specialized) are considered.

answered Jun 27 '13 at 13:28

Balog Pal

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Also if someone uses the function from within another template. – Dominik Grabiec Jun 27 '13 at 14:05
1

@Daemin: Can you expand on what you mean? – David Rodríguez - dribeas Jun 27 '13 at 14:18
In a function `template T foo(const T& a, const T& b)` they call `min(a, b);` hence it would use the templated version regardless of any non-templated overloads. – Dominik Grabiec Jun 27 '13 at 14:23

Template function specialization vs. overloading

1 Answers1