c++ - 使用 operator<< 重载打印派生对象不起作用

Question

为什么我不能使用这种方法打印派生对象？我怎样才能解决这个问题，以便派生打印输出打印“派生大小=8”。理想情况下，我希望将打印代码保留在类代码之外。

#include <iostream>

class base
{
public:
   base(int size) : size_(size) {}

   virtual int get_size() const { return size_; }

private:
   int size_;
};

std::ostream& operator<<(std::ostream& os, const base& obj) {
   os << "base size=" << obj.get_size() << std::endl;
   return os;
}

class derived1 : public base
{
public:
   derived1(int size) : base(size) {}

   virtual int get_size() const { 
      return 2 * base::get_size();
   }
};

std::ostream& operator<<(std::ostream& os, const derived1& obj) {
   os << "derived1 size=" << obj.get_size() << std::endl;
   return os;
}

int main(int argc, char* argv[]) {

   base* b1 = new base(3);
   std::cout << "b1 size is: " << b1->get_size() << std::endl;

   std::cout << *b1 << std::endl;

   base* b2 = new derived1(4);
   std::cout << "b2 size is: " << b2->get_size() << std::endl;

   std::cout << *b2 << std::endl;

   delete b1;
   delete b2;
   return 0;
}

Output:
b1 size is: 3
base size=3

b2 size is: 8
base size=8

更新:

我根据有效的 ghostsofstandardspast 进行了如下更改:

#include <iostream>

class base
{
public:
   base(int size) : size_(size) {}

   virtual int get_size() const { return size_; }
   friend std::ostream& operator<<(std::ostream &os, const base& obj);

private:
   virtual std::ostream &print(std::ostream &os) const {
      return os << "base size=" << get_size() << std::endl;
   }

   int size_;
};

std::ostream& operator<<(std::ostream& os, const base& obj) {
   obj.print(os);
   return os;
}

class derived1 : public base
{
public:
   derived1(int size) : base(size) {}

   virtual int get_size() const { 
      return 2 * base::get_size();
   }

   friend std::ostream& operator<<(std::ostream &os, const derived1& obj);
private:
   virtual std::ostream &print(std::ostream &os) const {
      return os << "derived1 size=" << get_size() << std::endl;
   }
};


int main(int argc, char* argv[]) {

   base* b1 = new base(3);
   std::cout << "b1 size is: " << b1->get_size() << std::endl;

   std::cout << *b1 << std::endl;

   base* b2 = new derived1(4);
   std::cout << "b2 size is: " << b2->get_size() << std::endl;

   std::cout << *b2 << std::endl;
   return 0;
}

Answer 1

operator<<不是虚成员函数，因此它不能像其他成员函数那样以多态方式使用。相反，您应该委托(delegate)给虚拟成员。

同时更改您的 operator<<重载到名为 print 的虚拟成员函数或类似的东西。然后，重载 operator<<委托(delegate)给那个:

std::ostream &operator<<(std::ostream &os, const Base &obj) {
    obj.print(os); //or whatever name you choose
    return os;
}

如果你想使成员函数私有(private)，使operator<<一个 friend 。

最后，请注意，打印指向您的类对象的指针只会打印地址。调用 operator<< ，您需要打印对象本身:

Base *b = /*whatever*/;
std::cout << *b;

---

此技术用于多态 operator<<在 this cppquiz question 中展示.我强烈建议您完成这些问题。