std::enable_if
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| Definido no cabeçalho <type_traits>
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| template< bool B, class T = void > struct enable_if; |
(desde C++11) | |
Se
B é true, std::enable_if tem um type membro pública typedef, igual a T, caso contrário, não há nenhum membro 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.
Este metafunção é usado para remover condicionalmente funções e classes de resolução de sobrecarga com base em características de tipo e para fornecer sobrecargas de função separados e especializações para características de tipo diferente. std::enable_if pode ser usado como um argumento de função adicional (não aplicável a sobrecargas de operador), como um tipo de retorno (não aplicável a construtores e destruidores), ou como um modelo de classe ou parâmetro do modelo função.
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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Índice |
[editar] Tipos de membro
| 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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ou
T ou não esse membro, de acordo com o 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] Possível implementação
template<bool B, class T = void> struct enable_if {}; template<class T> struct enable_if<true, T> { typedef T type; }; |
[editar] Exemplo
#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 }
Saída:
foo1: float foo1: int
