C++ 复制赋值运算符行为

Question

我使用下面的代码来测试复制赋值运算符的行为：

#include <iostream>
using namespace std;

int group_number = 10; // Global

class Player {
public:
  explicit Player(const int &g): group(g)
  {
  }

  Player& operator=(const Player& rhs)
  {
      cout << "Copy assignment operator called " << endl;
      return *this;
  }

  void set_stamina(int i) {stamina = i; }
  int  get_stamina()      {return stamina; }
  const int &get_group()  const { return group; }

protected:
  const int &group;

private:
  int stamina;
}; // End of Player class

class Round {
public:
  Round(const Player &p)
  {
    main_player = &p;
  }

  const Player &get_main_player() const
  {
    return *main_player;
  }

protected:
  const Player *main_player;
}; // End of Round class

// Test code
int main()
{
  Player p1(group_number);
  Round r(p1);
  p1.set_stamina(100);

  // Add player2 in the same group as p1
  Player p2(group_number);
  p2 = r.get_main_player();
  cout << "This is p1's stamina: "    << p1.get_stamina()
       << ", and p1's group number: " << p1.get_group() << endl;;
  cout << "This is p2's stamina: "    << p2.get_stamina()
       << ", and p2's group number: " << p2.get_group() << endl;;

  return 0;
}

我预计 p1 和 p2 具有相同的值 stamina。但是输出显示 p2 的 stamina 与 p1 的

不同

Copy assignment operator called
This is p1's stamina: 100, and p1's group number: 10
This is p2's stamina: 241098768, and p2's group number: 10

为什么复制赋值运算符没有将 p1 的 stamina 的值复制到 p2？

Answer 1

Why the copy assignment operator not copying the value of p1's stamina to p2?

您定义了运算符。您没有在运算符的定义中复制任何值，因此不会复制任何值。

您可以这样分配成员：

stamina = rhs.stamina;

Answer 2

您没有在复制赋值运算符中复制数据成员stamina。因此，只需在复制赋值运算符中添加 stamina = rhs.stamina; 即可获得预期的输出。所以 operator= 的修改定义看起来像：

Player& operator=(const Player& rhs)
  {
      cout << "Copy assignment operator called " << endl;
      stamina = rhs.stamina; //ADDED THIS 
      return *this;
  }

另外请注意，在您的原始代码片段中，由于数据成员 stamina 未初始化，因此它具有 不确定值 。使用这个未初始化的变量（你在写 p2.get_stamina(); 时所做的）是 未定义的行为。

Undefined behavior means anything¹ can happen including but not limited to the program giving your expected output. But never rely(or make conclusions based) on the output of a program that has undefined behavior.

所以您看到的输出是未定义行为的结果。正如我所说，不要依赖具有 UB 的程序的输出。

因此，使程序正确的第一步是删除 UB。 然后并且只有那时你可以开始对程序的输出进行推理。

为了解决这个更严重的问题只需对数据成员stamina使用in-class initilaizer。

class Player {
    //other members here
private:
  int stamina = 0; //USE IN-CLASS INITIALIZER
};

^{¹有关未定义行为的技术上更准确的定义，请参阅 this 其中提到：没有对程序行为的限制.}

C++ 复制赋值运算符行为

C++ copy assignment operator behaviour

c++

class

assignment-operator

copy-assignment

explicit-constructor