我是否应该期望单一继承中的上下转换不会调整指针? [英] Should I expect that upcasts and downcasts in single inheritance don't adjust the pointer?

问题描述

假设我有:

class Base {
public:
    virtual void Nothing() {}
};
class MiddleDerived : public Base {
    virtual void Nothing() {}
};
class Derived : public MiddleDerived {
    virtual void Nothing() {}
};

我的代码如下:

Derived* object = new Derived();
Base* base = object; //implicit conversion here

void* derivedVoid = object;
void* baseVoid = base;

我应该期待baseVoid == derivedVoid吗?

我知道大多数实现都是以这种方式工作的，但是可以保证吗?

I know that most implementations work this way but is it guaranteed?

推荐答案

我认为处理此问题的标准部分为§5.2.9(13)

I think the part of the standard that deals with this is §5.2.9 (13)

(简而言之)声明T *转换为void *然后返回T *应当引用同一对象.

Which states (in a nutshell) that a T* cast to a void* and then back to a T* shall refer to the same object.

但是没有规定Derived的地址必须与它的Base的地址相同.

However there is no stipulation The address of Derived must be the same as the address of its Base.

所以我的回答是:不-期望这种等效的代码是格式错误的，它会引起不确定的行为".

So my answer would be, "no - code that expects this equivalency is ill-formed inviting undefined behaviour".

这篇关于我是否应该期望单一继承中的上下转换不会调整指针?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！