std::enable_if
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| Defined in header <type_traits>
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| template< bool B, class T = void > struct enable_if; |
(ya que C + +11) | |
Si
B es true, std::enable_if tiene un type miembro público typedef, igual a T, de lo contrario, no hay ningún miembro typedef . Original:
If
B is true, std::enable_if has a public member typedef type, equal to T; otherwise, there is no member typedef. The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Esta metafunción se utiliza para eliminar condicionalmente funciones y clases de resolución de sobrecarga basada en caracteres de tipo y proporcionar sobrecargas separar la función y especializaciones para los rasgos de tipos. std::enable_if puede ser utilizado como un argumento de la función adicional (no aplicable a sobrecargas de operadores), como un tipo de retorno (no aplicable a los constructores y destructores), o como una plantilla de clase o parámetro de plantilla de función .
Original:
This metafunction is used to conditionally remove functions and classes from overload resolution based on type traits and to provide separate function overloads and specializations for different type traits. std::enable_if can be used as an additional function argument (not applicable to operator overloads), as a return type (not applicable to constructors and destructors), or as a class template or function template parameter.
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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Contenido |
[editar] Tipos de miembros
| Tipo
Original: Type The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
Definition |
type
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ya sea
T o no dicho miembro, en función del valor de BOriginal: either T or no such member, depending on the value of BThe text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
[editar] Posible implementación
template<bool B, class T = void> struct enable_if {}; template<class T> struct enable_if<true, T> { typedef T type; }; |
[editar] Ejemplo
#include <type_traits> #include <iostream> // foo1 overloads are enabled via the return type template<class T> typename std::enable_if<std::is_floating_point<T>::value, T>::type foo1(T t) { std::cout << "foo1: float\n"; return t; } template<class T> typename std::enable_if<std::is_integral<T>::value, T>::type foo1(T t) { std::cout << "foo1: int\n"; return t; } // foo2 overload is enabled via a parameter template<class T> T foo2(T t, typename std::enable_if<std::is_integral<T>::value >::type* = 0) { return t; } // foo3 overload is enabled via a template parameter template<class T , class = typename std::enable_if<std::is_integral<T>::value>::type > T foo3(T t) // note, function signature is unmodified { return t; } // A is enabled via a template parameter template<class T, class Enable = void> class A; // undefined template<class T> class A<T, typename std::enable_if<std::is_floating_point<T>::value >::type> { }; int main() { foo1(1.2); // OK, calls the first version of foo1() foo1(10); // OK, calls the second version of foo1() // foo2(0.1); // compile-time error foo2(7); // OK // A<int> a1; // compile-time error A<double> a1; // OK }
Output:
foo1: float foo1: int