以下前向声明的多继承指针转换代码如何工作？

发布于 2024-08-19 13:53:04 字数 419 浏览 3 评论 0原文

在下面的代码中，指针如何转换&多重继承一起玩？

class Foo {
  public:
  virtual void someFunc();
};

class Bar;


void someWork(Bar *bar) {
  ((Foo*) bar)->someFunc();
}

class Bar: public Zed, public Foo {
...
virtual void someFunc() { ... do something else ... }
}

Bar bar;

int main() {
  someWork(&bar);
}

我的理解有点动摇。

一方面，someWork 对 Bar 一无所知，所以这不应该起作用；但另一方面，我已经向前宣布了吧。

谢谢！

原文

In the followint code, how does the pointer conversion & multi-inheritance play together?

class Foo {
  public:
  virtual void someFunc();
};

class Bar;


void someWork(Bar *bar) {
  ((Foo*) bar)->someFunc();
}

class Bar: public Zed, public Foo {
...
virtual void someFunc() { ... do something else ... }
}

Bar bar;

int main() {
  someWork(&bar);
}

My understanding is kinda shaky.

On one hand, someWork knows nothing about Bar, so this shouldn't work; but on the other hand, I have forward declared Bar.

Thanks!

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怼怹恏 2024-08-26 13:53:04

这是行不通的，而且它的作用也不完全是你想象的那样。您对 c 风格的强制转换的使用

(Foo*) bar

在这种情况下，是不正确的。您想要做的是将 Bar* 向上转换为 Foo* （即，从指向派生类的指针执行 static_cast指向基类的指针）。

然而，由于此时 Bar 的定义不可用，编译器不知道 Foo 是 Bar 的基类。因此，static_cast 失败，编译器回退并使用 reinterpret_cast，这根本不是一回事。

This doesn't work and it isn't doing quite what you think it is. Your use of the c-style cast:

(Foo*) bar

is incorrect in this case. What you are trying to do is upcast the Bar* to a Foo* (i.e., perform a static_cast from a pointer to a dervied class to a pointer to a base class).

Since the definition of Bar is not available at this point, however, the compiler does not know that Foo is a base class of Bar. Thus, the static_cast fails and the compiler falls back and uses a reinterpret_cast, which is not at all the same thing.

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