为什么C风格转换允许你转换为私有基类？ [英] Why does C style cast allow you to convert to a private base class?

问题描述

说我们有这个代码

class A {
public:
    A() : x(1) {}
    virtual ~A() {}

    int x;
};

class B {
public:
    B() : y(2) {}
    virtual ~B() {}

    void g()
    {
        cout << "B::" << y << endl;
    }

    int y;
};

class C : private A, private B {
public:
    void f()
    {
        B* p = static_cast<B*>( this );
        p->g();
    }
};

int main()
{
    C c;
    ((B*)&c)->g();

    return 0;
}

主函数中的C风格转型无法正确表达C ++ casts（ static_cast ， dynamic_cast ， reinterpret_cast ）。但是，首先允许这样做的原因是什么？它不会伤害封装吗？

The C style cast in the main function cannot be correctly expressed in terms of the C++ casts (static_cast, dynamic_cast, reinterpret_cast). But what is the reason to allow this in the first place? Doesn't it hurt encapsulation?

UPDATE
这不是重复的链接问题，因为这个问题是关于设计C ++中的决策。它不问我能用或不能用语言做什么，它问为什么可能做出某些决定。

UPDATE This is not a duplicate of the linked question, because this question is about design decisions in C++. It does not ask what I can or cannot do with the language, it asks why certain decisions might have been made.

推荐答案

因为在C中允许使用这个转换将任何指针转换为任何其他指针，并且C ++尽可能地是C兼容的，但尝试做一个好的工作是正确的，当涉及到类，所以C风格的转型在这种情况下比 reinterpret_cast 更强。

It's because in C it was allowed to convert any pointer to any other pointer using this cast and C++ tries to be C-compatible as much as possible, but tries to do a good job to be correct when it comes to classes, so C style cast is stronger than reinterpret_cast in this situation.

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