C ++：比较基类和派生类的指针

问题描述

| 我希望在诸如此类的情况下比较指针时有关最佳做法的一些信息：

class Base {
};

class Derived
    : public Base {
};

Derived* d = new Derived;
Base* b = dynamic_cast<Base*>(d);

// When comparing the two pointers should I cast them
// to the same type or does it not even matter?
bool theSame = b == d;
// Or,bool theSame = dynamic_cast<Derived*>(b) == d?

    

解决方法

如果要比较任意的类层次结构，安全的选择是使它们多态并使用

dynamic_cast

class Base {
  virtual ~Base() { }
};

class Derived
    : public Base {
};

Derived* d = new Derived;
Base* b = dynamic_cast<Base*>(d);

// When comparing the two pointers should I cast them
// to the same type or does it not even matter?
bool theSame = dynamic_cast<void*>(b) == dynamic_cast<void*>(d);

考虑到有时您不能使用static_cast或从派生类到基类的隐式转换：

struct A { };
struct B : A { };
struct C : A { };
struct D : B,C { };

A * a = ...;
D * d = ...;

/* static casting A to D would fail,because there are multiple A\'s for one D */
/* dynamic_cast<void*>(a) magically converts your a to the D pointer,no matter
 * what of the two A it points to.
 */

如果

A

是虚拟继承的，您也不能static_cast到

D

。     ,在上述情况下，您不需要任何强制转换，只需简单的

Base* b = d;

就可以。然后，您可以像现在进行比较一样比较指针。