类型作为returntype在c ++ [英] type as returntype in c++

问题描述

有可能从函数
中返回一个类型作为returntype，并使用类似这样的成员变量：

Is there a possibility to return a type as a returntype from a function and use it for a member variable using something like this:

constexpr type myFunction(int a, int b){
     if(a + b == 8) return int_8t;
     if(a + b == 16) return int_16t;
     if(a + b == 32) return int_32t;
     return int_64t;
}

template<int x, int y>
class test{
    using type = typename myFunction(x, y);
    private:
    type m_variable;
};

当在Qt中尝试这个例子时，它说

When trying this example in Qt it says

Error: 'constexpr' does not name a type
Error: 'test' is not a template type
class test{
      ^

在先前的问题中，有人向我显示 http://en.cppreference.com/w/cpp/types/conditional 这个函数，但它只适用于2种类型。

In an earlier question someone showed me http://en.cppreference.com/w/cpp/types/conditional this function, but it works only for 2 Types.

推荐答案

你不能用普通的函数做到这一点。然而，它很容易使用模板元编程。这种模板有时称为类型函数。

You cannot do this with a normal function. However, it is easily done using template meta-programming. This kind of template is sometimes called a type function.

#include <cstdint>

template<int bits> struct integer { /* empty */ };

// Specialize for the bit widths we want.
template<> struct integer<8>  { typedef int8_t  type; };
template<> struct integer<16> { typedef int16_t type; };
template<> struct integer<32> { typedef int32_t type; };

可以这样使用。

using integer_type = integer<16>::type;
integer_type number = 42;

请记住整数< T> :: type c> typename 关键字if T 本身是一个模板参数。

Remember to precede integer<T>::type with the typename keyword if T is itself a template parameter.

我把它作为一个练习，扩展到一个模板，接受两个整数作为参数，并返回基于两者之和的合适的类型。

I leave it as an exercise to you to extend this to a template that accepts two integers as parameters and returns the appropriate type based on the sum of the two.

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