Why does std::is_same give a different result for the two types?

Question

In the code below, why do the two ways of invoking fun: fun(num) and fun<const int>(num), give a different result when compiling?

#include <iostream>
using namespace std;

template<typename T, typename = typename enable_if<!std::is_same<int, T>::value>::type>
void fun(T val)
{
    cout << val << endl;
}

int main(void)
{
    const int num = 42;
    fun(num);  //ERROR!

    fun<const int>(num);  //Right

    return 0;
}

Madelaine

An int prvalue is never const qualified. – Fronde

Answer 1

The parameter is declared as pass-by-value; then in template argument deduction, the top-level const qualifier of the argument is ignored.

Before deduction begins, the following adjustments to P and A are made:

1) If P is not a reference type,

a) ...

b) ...

c) otherwise, if A is a cv-qualified type, the top-level cv-qualifiers are ignored for deduction:

So given fun(num), the template parameter T will be deduced as int, not const int.

If you change the parameter to pass-by-reference, the const part will be preserved. e.g.

template<typename T, typename = typename enable_if<!std::is_same<int, T>::value>::type>
void fun(T& val)

Then for fun(num), T will be deduced as const int.

Wildawildcat

What's more, the type of num is const int, the type of the value of num is just int. There is no such thing a value (a prvalue, technically) of type const int. – Fronde