#include <iostream>
#include <cmath>
#include <type_traits>
template <typename T>
void f(T, T) // 1
{
std::cout << "Primary\n";
}
template <typename T>
void f(T, std::enable_if_t<std::is_floating_point_v<T>, T>) // 2
{
std::cout << "Special\n";
}
/*template <typename T>
std::enable_if_t<std::is_floating_point_v<T>> f(T, T) // 3
{
std::cout << "Special\n";
}*/
int main()
{
f(1.1, 1.1); // prints 'Primary'
return 0;
}
In the above code, std::enable_if is applied to the function type of the second function template overload. The function is called with [T = double] deduced, and it calls overload 1. However, if I comment out overload 2 and replace it with overload 3, then the compiler will complain that the call is ambiguous. I expected that in the first case too, why does the compiler prefer overload 1 over 2?
I read the section on 'function template overloading', but to me it looks like overload 2 is more specialized.
