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;
}

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. eg

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 .