C ++指针多继承乐趣 [英] C++ pointer multi-inheritance fun

问题描述

我写一些代码涉及从基本的计数指针类继承;和一些复杂的C ++弹出。我已将其减少如下：

假设我有：

  class A {}; 
 class B {}; 
 class C：public A，public B {}; 
 
 C c; 
 C * pc =& c; 
 B * pb =& c; 
 A * pa =& c 
 
 //指向一个有效的A对象？ 
 // pb指向一个有效的B对象？ 
 
 // pa = pb？ 
  

此外，还有：

  // pc ==（C *）pa？ 
 // pc ==（C *）pb？ 
  

$ b

谢谢！

解决方案<

pb指向有效的B对象？

/ li>

是的， C * 以便 pa 和 pb 指向正确的地址。

< blockquote>

• 是pa == pb吗？

不，通常不是。在同一地址不能有 A 对象和 B 对象。

此外，

• pc ==（C *）pa？


• pc ==（C *）pb？

该转换将指针转换回 C 对象的地址，因此两个等式都为真。

I'm writing some code involving inheritance from a basic ref-counting pointer class; and some intricacies of C++ popped up. I've reduced it as follows:

Suppose I have:

class A{};
class B{};
class C: public A, public B {};

C c;
C* pc = &c;
B* pb = &c;
A* pa = &c;

// does pa point to a valid A object?
// does pb point to a valid B object?

// does pa == pb ?

Furthermore, does:

// pc == (C*) pa ?
// pc == (C*) pb ?

Thanks!

解决方案

• does pa point to a valid A object?
• does pb point to a valid B object?

Yes, the C* gets converted so that pa and pb point to the correct addresses.

• does pa == pb ?

No, usually not. There can't be an A object and a B object at the same address.

Furthermore, does

• pc == (C*) pa ?
• pc == (C*) pb ?

The cast converts the pointers back to the address of the C object, so both equalities are true.

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