通过转换基类指针设置派生类字段 [英] Set derived class field by converting base class pointer

问题描述

class A
{
public: 
    int a;
};
class B:public A
{
public:
    int b;
    void foo()
    {
        b=a*a;
    }
};
int _tmain(int argc, _TCHAR* argv[])
{ 
    A * a=new A;
    a->a=10;
    ((B*)a)->foo();
    cout<<((B*)a)->b;
}

它适用于 b = 100 ，但我不知道它的工作原理。 b 存储在哪里？我只是不知道它的调用到谷歌它。

It's working for b=100, but I dont know by which rules it works. Where is b stored? I just don't know how its called to google it.

推荐答案

基本上，这里发生的是未定义的行为。它没有特殊的名字;很可能是被称为编程错误。您的类 A 的内存布局是：

Basically, what is happening here is undefined behaviour. It doesn't have a special name; most likely it is called a programming mistake. The memory layout of your class A is:

int a;

B 的内存布局是： / p>

The memory layout of B is:

int a;
int b;

所以在你的情况下，你只需为 a 但是你很幸运，空间立即之后它是免费的（以便没有其他信息被覆盖），并且它没有边界在未分配的空间（否则，当尝试写入未分配的页面时可能会发生故障） 。因此， b 存储在可用空间中。

So in your case, you only allocate space for a but you are lucky that the space immediately after it is free (so that no other information is overwritten) and that it doesn't border on unallocated space (otherwise, a fault might occur when trying to write to an unallocated page). So b is stored in free space.

总之：工作！

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