Question

class Musician {
public: virtual void play() { std::cout << "Play an instrument"; }
 };

class Guitarist : public Musician {
public: void play() { std::cout << "Play a guitar"<<endl; }
};

int main(){
Musician armstrong;
Guitarist steve;
Musician* m = &armstrong;
m->play();
m = &steve; //type of (m)=Musician type of (*m) = Guitarist???
}

为什么 (*m) = Guitarist 的类型？为什么如果我在 Musician 类的函数中删除 virtual 关键字，然后类型为 (*m) = Musician？

class Musician {
public: void play() { std::cout << "Play an instrument"; } //this makes type of(*m) = Musician
 };

如果我在最后添加 armstrong = static_cast<Musician>(steve);，它不会给出错误，但不会导致 armstrong 的任何更改（地址和类型）。那条线是什么意思？而 steve = static_cast<Guitarist>(armstrong); 给出错误：

 error: no matching function for call to 'Guitarist::Guitarist(Musician&)'
   27 | steve=static_cast<Guitarist>(armstrong);
      |                                       ^

Answer 1

我在这里学习时曾尝试过这样做，我提出了一个解决方案，您可以将 A 替换为 Musician，将 B 替换为 Guitarist。

#include <iostream>

class A {
public:
  void Same(){
      std::cout << "From A" << std::endl;
  }
  virtual A& operator= (const A& a) {
    assign(a);
    return *this;
  }

protected:
  void assign(const A& a) {
  }
};

class B : public A {
public:
  void Same(){
      std::cout << "From B" << std::endl;
  }
  virtual B& operator= (const A& a) {
    if (const B* b = dynamic_cast<const B*>(&a))
      assign(*b);
    return *this;
  }

protected:
  void assign(const B& b) {
    A::assign(b); 
  }
};

int main(){
    A a;
    B b;
    a.Same();
    a = b;
    a.Same();
    return 0;
}

类、继承和虚函数

1 个答案: