使用inhertiance和polymorphism重载'<<'？

Question

以下是代码外观的粗略示例,问题是如何让DerivedOne和DerivedTwo具有重载的<<运算符,但将这些对象存储在Base*的向量中.

至于我想要达到的目标; 我希望能够遍历对象向量并输出我在DerivedOne和DerivedTwo中告诉它的信息.

vector<Base*> objects;

class Base
{
 private:
 object Data
 public:
 object getData() { return Data; }
};

class DerivedOne : public Base
{
}

class DerivedTwo : public Base
{
}

现在我知道有这个,但它不适用于我的目的.

friend ostream &operator<<(ostream &stream, Object object)
{
    return stream << "Test" << endl;
}

Answer 1

使您的虚方法成为私有,以便将对象的使用方式与派生类的行为自定义方式分开.

这与其他答案解决方案类似,但虚拟方法是私有的:

#include <iostream>

namespace {
  class Base {
    // private (there is no need to call it in subclasses)
    virtual std::ostream& doprint(std::ostream&) const = 0;
  public:
    friend std::ostream& operator << (std::ostream& os, const Base& b) {
      return b.doprint(os); // polymorphic print via reference
    }

    virtual ~Base() {} // allow polymorphic delete
  };


  class DerivedOne : public Base {
    std::ostream& doprint(std::ostream& os) const {
      return os << "One";
    }
  public:
    DerivedOne() { std::cerr << "hi " << *this << "\n"; } // use << inside class
    ~DerivedOne() { std::cerr << "bye " << *this << "\n"; }
  };
}

例

#include <memory>
#include <vector>

int main () {
  using namespace std;
  // wrap Base* with shared_ptr<> to put it in a vector
  vector<shared_ptr<Base>> v{ make_shared<DerivedOne>() };
  for (auto i: v) cout << *i << " ";
  cout << endl;
}

产量

hi One
One 
bye One

Answer 2

你可以这样做:

struct Base {
    virtual ~Base () {}
    virtual std::ostream & output (std::ostream &) const = 0;
};

std::ostream & operator << (std::ostream &os, const Base &b) {
    return b.output(os);
}

然后,当应用operator<<on时Base,它将调用派生的输出方法.