c++ - 从 void* 到基类指针的转换

Question

我有一些层次结构:基类、派生类和一些将用户数据存储为 void* 的结构。该 void 可以存储 Base 和 Derived 类指针。主要问题是我不知道那里存储的是基指针还是派生指针。

class Base
{
public:
  int type;
};
class Derived: public Base
{};<p></p>

<p>Base* base;//init base pointer
Derived* derived;//init derived pointer
void* base_v = base;
void* derived_v = derived;
//void pointers are correct. They point to base and derived variables.</p>

<p>//try to get type field after converting pointers back
Derived* d_restored = (Derived*)derived_v;//d_restored correct
Base* b_restored = (Base*)base_v;//b_restored correct
Base* d_restored_to_base = (Base*)derived_v;// INCORRECT
</p>

如何将 void* 转换为两个指针的 [type] 字段？提前致谢。

Answer 1

void* 只能转换回其原始类型。当您将 Derived* 存储在 void* 中时，您只能转换回 Derived*，不能 基础*。

这在多重继承中尤为明显，因为您的派生对象不一定与您的基地址位于同一地址。如果你真的需要用 void* 存储东西(和检索东西)，总是首先转换为基本类型，这样你就有了一个稳定的方法来取回对象:

#include <iostream>

struct base { int type; };
struct intruder { int iminyourclassstealingyourbase; };
struct derived : intruder, base {};

int main()
{
    derived d; d.type = 5;

    void* good = (base*)&d;
    void* bad = &d;

    base* b1 = (base*)good;
    base* b2 = (base*)bad;

    std::cout << "good: " << b1->type << "\n";
    std::cout << "bad: " << b2->type << "\n";
}

如果您随后想返回派生类型，请使用 dynamic_cast(或 static_cast，如果您保证它必须是派生类型。)