C++：比较基类和派生类的指针

Question

我想要一些有关在类似这样的情况下比较指针时的最佳实践的信息：

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?

Answer 1

如果您想比较任意类层次结构，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 为 D 。

Answer 2

在上述情况下，您不需要任何转换，简单的 Base* b = d; 就可以了。然后您可以像现在比较一样比较指针。