Implicit conversion from true value to std::true_type

Question

This is a simple template progs that I wrote to study C++:

#include <type_traits>
#include <iostream>

using namespace std;

template<typename T>
T foo(T t, true_type)
{
    cout << t << " is integral! ";
    return 2 * t;
}


template<typename T>
T foo(T t, false_type)
{
    cout << t << " ain't integral! ";
    return -1 * (int)t;
}

template<typename T>
T do_foo(T t){
    return foo(t, is_integral<T>());
}

int main()
{
    cout << do_foo<int>(3) << endl;
    cout << do_foo<float>(2.5) << endl;
}

It doesn't do anything to fancy, but it does compile and work.

I am wondering how does the part is_integral<T>() work?

I was reading this : http://en.cppreference.com/w/cpp/types/is_integral and I can't find any specific description of this behavior - no definition of operator()

Answer 1

is_integral<T> is a type that inherits either from true_type or false_type.

is_integral<T>() is a constructor call, so that an instance of one of those types is an argument to the call to foo. The overload is then selected according to which one it is.