如何删除相似的经过ref限定的成员函数之间的代码重复？ [英] How do I remove code duplication between similar ref-qualified member functions?

问题描述

类似于如何删除相似的const和非const成员函数之间的代码重复？，我要删除

假设我有一个类似这样的类：

Let's say I have a class that's something like this:

class MyStringBuilder
{
    std::string member;
public:
    // Other functions
    std::string create() const& {
        // Some work
        std::string result = member;
        // More work
        return result;
    }

    std::string create() && {
        // Some work
        std::string result = std::move(member);
        // More work
        return result;
    }
};

对于构建器对象我们想这样做并非不可想象，因为如果我们使用 MyStringBuilder 完成。

It's not inconceivable that we would want to do this for a builder object, as it saves a copy if we are done with the MyStringBuilder.

除了使用成员的地方，<$ c $之间的代码c> const& 版本与& 版本相同。这两个函数之间的唯一区别是&& 版本 std :: move 时，任何成员

Except for where members are used, the code between the const& version and && version are identical. The only difference between the two functions is that the && version std::moves any members whenever they are referenced.

如何避免此代码重复？

推荐答案

您可以做的一件事是，您可以在非成员函数中实现逻辑，并将 * this 的类型作为模板参数：

One thing you can do is you can implement the logic in a non-member function and take the type of *this as a template parameter:

class MyStringBuilder
{
    std::string member;

    template<typename Self>
    static std::string create_impl(Self&& self) {
        // Some work
        std::string result = std::forward<Self>(self).member;
        // More work
        return result;
    }
public:
    // Other functions
    std::string create() const& {
        return create_impl(*this);
    }

    std::string create() && {
        return create_impl(std::move(*this));
    }
};

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