如何转换为私下派生的子类型？ [英] How to cast to privately derived child type?

问题描述

以下是尝试使用 operator == 的修改语义实现共享指针：

The following is an attempt at implementing a shared pointer with a modified semantics of operator==:

template <typename T>
struct deref_shared_ptr: private std::shared_ptr<T> {
    using Base = std::shared_ptr<T>;
    // ... using statements to include functionality from the base.

    bool operator==(const deref_shared_ptr rhs) const {
        return (**this == *rhs);
    }
};

我正在努力实现一个等效的 std :: make_shared 这种类型。这是我的尝试：

I am struggling with implementing an equivalent of std::make_shared for this type. This is my attempt:

template< class T, class... Args >
deref_shared_ptr<T> make_deref_shared( Args&&... args ) {
    return reinterpret_cast<deref_shared_ptr<T>>(std::make_shared<T>(args...));
}

这不起作用：编译器（ g ++ 5.4 .0 ）投诉无效投射。为什么它不工作，我应该做什么，而不是这个演员？

This does not work: the compiler (g++ 5.4.0) complains about an invalid cast. Why does it not work and what should I do instead of this cast?

推荐答案

您看到此编译器错误消息，因为 reinterpret_cast 通过私人继承。请检查以下主题有关此主题： c ++ casts之间的区别，转换只能通过c风格转换处理。

You see this compiler error message because the reinterpret_cast cannot make casts through the private inheritance. Please check the following themes on this topic: difference between c++ casts, conversion which may be handled by c-style cast only.

只有通过 private 继承是c风格的转换。因此，更改您的示例如下使您的示例工作：

The only way to go through the private inheritance is the c-style cast. So, changing your example as follows makes your example work:

template< class T, class... Args >
deref_shared_ptr<T> make_deref_shared(Args&&... args) {
    return (deref_shared_ptr<T>)(std::make_shared<T>(args...));
}

c风格的转换在一般情况下是不安全的，因为它可能工作不正确的情况下多重遗产和一些其他情况下，但AFAIK它是安全的在这种情况下。

The c-style cast is not safe in the general case since it may work incorrectly in cases of multiple inheritance and some other cases, but AFAIK it's safe in this case.

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