如何使用带模板的类成员的尾随返回类型 [英] How to use trailing return type with a templated class member

问题描述

我正在尝试实现以下类:

I'm trying to implement the following class:

template <typename Container>
class reverse_adaptor
{
public: // Construction
    reverse_adaptor(Container &container) :
        m_container(container)
    {}

public: // STL container static polymorphism
    auto begin() const -> decltype(m_container.rbegin())
    {
        return m_container.rbegin();
    }

    auto end() const -> decltype(m_container.rend())
    {
        return m_container.rend();
    }

private: // Members
    Container &m_container;
};

我使用尾随返回类型的原因是因为我不知道m_container是否为const，所以我让编译器为我解决.但是，出现以下编译器错误:

The reason I'm using the trailing return type is because I don't know if m_container will be const or not, so I let the compiler work it out for me. However, I'm getting the following compiler errors:

/Users/mark/Blah/Stdx.h:77:40:在其中没有名为"m_container"的成员 'stdx :: reverse_adaptor>>'

/Users/mark/Blah/Stdx.h:77:40: No member named 'm_container' in 'stdx::reverse_adaptor > >'

我认为这可能与模板类型的多阶段传递有关，因此将其更改为decltype(this->m_container.rbegin())，但这也不起作用.

I thought it might be related to the multi-stage pass of templated types, so changed it to read decltype(this->m_container.rbegin()), but that didn't work either.

如何使它正常工作?

示例- http://ideone.com/ekVYlH

推荐答案

函数的跟踪返回类型是其"签名"(声明)的一部分，而不是其"body"(定义)，因此，它只能看到之前声明的名称.

The trailing-return-type of a function is part of its "signature" (declaration), not of its "body" (definition), and as such, it only sees names that were declared before.

在声明您的begin成员函数时，尚未声明m_container. (请注意，该问题并不特定于模板类.)

At the point where you declare your begin member function, m_container hasn't been declared yet. (Note that the issue is not specific to template classes.)

• 您可以将m_container的声明在类定义中上移(但是这会迫使您将私有成员放在公共接口之前，这与通常的做法相反……).

• You could move the declaration of m_container up in the class definition (but it forces you to put private members before public interface, which is contrary to common practice...).

您可以使用 declval 解决:将m_container替换为<decltype中的c6>: http://ideone.com/aQ8Apa

You can work-around with declval: replace m_container with std::declval<Container&>() inside the decltype: http://ideone.com/aQ8Apa

(如评论中所述，在C ++ 14中，您可以删除尾随的返回类型，而只需将decltype(auto)用作常规"返回类型.)

(As said in the comments, in C++14 you'll be able to drop the trailing return type and just use decltype(auto) as the "normal" return type.)

附录:至于为什么可以在成员函数的类内主体中使用尚未声明的成员，但不能在尾随返回类型中使用的原因，原因如下:

Addendum: As for why you can use not-yet-declared members inside the in-class body of a member function but not in the trailing return type, it's because the following:

class Foo {
public:
    auto data() const -> decltype(m_data)
    {
        return m_data;
    }
private:
    SomeType m_data;
};

将是" _{[免责声明:非正式用语！"} ，类似于由编译器重写"的形式:

will be _{[disclaimer: informal wording!]} kind of "rewritten by the compiler" into something equivalent to this:

class Foo {
public:
    inline auto data() const -> decltype(m_data); // declaration only
private:
    SomeType m_data;
};

// at this point, all members have been declared

inline auto Foo::data() const -> decltype(m_data) // definition
{
    return m_data;
}

其中名称不能在声明之前使用(第一个decltype(m_data)违反).

where a name cannot be used before its declaration (which the first decltype(m_data) violates).

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