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:
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template<bool Cond, class T = void> struct enable_if {};
template<class T> struct enable_if<true, T> { typedef T type; };
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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
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// 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;
}
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Output:
i is odd: true
i is even: false
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