如何在C ++ 11中使用decltype引用当前类？ [英] How to refer current class using decltype in C++11?

问题描述

当我声明一个类静态方法时，是否可以使用 decltype （或任何其他类似样式）引用当前类？例如，

When I declare a class static method, is it possible to refer current class using decltype (or in any other similar style)? For example,

class
AAA
{
    static AAA const make();
};

我正在尝试制作类似的东西。

I am trying to make something like this.

class
AAA
{
    static decltype(*this) const make();  // Not working because there's no `this`.
};

* this 用于描述我想要做。我想知道一些 decltype（）表达式，可以将其解析为 AAA 。

The *this is used to describe what I want to do. I want to know some decltype() expression which can be resolved to AAA.

如果可能的话我该怎么做？

If it's possible how can I do that?

推荐答案

在C ++ 1y中，您可以这样做：

In C++1y you could do this:

class
AAA
{
public:
    static auto make()
    {
        return AAA();
    }
};

int main()
{
    AAA aaa = AAA::make();
}

在C ++ 11中这是不合法的，因为您需要指定返回值 make（）的类型。在C ++ 98/03/11中，您可以：

This is not legal in C++11 as you need to specify a return type for make(). In C++98/03/11 you can:

class
AAA
{
public:
    typedef AAA Self;

    static Self make()
    {
        return AAA();
    }
};

技术含量低，但可读性强。

That is low-tech, but very readable.

< aside>

您应避免按值返回const限定类型。这抑制了有效的移动语义。如果要避免分配给右值，则创建一个用& 限定的赋值运算符。

You should avoid returning const-qualified types by value. This inhibits efficient move semantics. If you want to avoid assigning to rvalues, then create an assignment operator qualified with &.

< / aside>

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