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Showing content from https://cplusplus.com/reference/type_traits/enable_if/ below:

class template

<type_traits>

std::enable_if

template <bool Cond, class T = void> struct enable_if;

Enable type if condition is met

The type T is enabled as member type enable_if::type if Cond is true.

Otherwise, enable_if::type is not defined.

This is useful to hide signatures on compile time when a particular condition is not met, since in this case, the member enable_if::type will not be defined and attempting to compile using it should fail.

It is defined with a behavior equivalent to:

1
2

template<bool Cond, class T = void> struct enable_if {};
template<class T> struct enable_if<true, T> { typedef T type; };

Template parameters

Cond: A compile-time constant of type bool.
T: A type.

Member types member type definition type T
(defined only if Cond is true)
Example

// enable_if example: two ways of using enable_if
#include <iostream>
#include <type_traits>

// 1. the return type (bool) is only valid if T is an integral type:
template <class T>
typename std::enable_if<std::is_integral<T>::value,bool>::type
  is_odd (T i) {return bool(i%2);}

// 2. the second template argument is only valid if T is an integral type:
template < class T,
           class = typename std::enable_if<std::is_integral<T>::value>::type>
bool is_even (T i) {return !bool(i%2);}

int main() {

  short int i = 1;    // code does not compile if type of i is not integral

  std::cout << std::boolalpha;
  std::cout << "i is odd: " << is_odd(i) << std::endl;
  std::cout << "i is even: " << is_even(i) << std::endl;

  return 0;
}

Output:

i is odd: true
i is even: false