使用自定义构造函数（C+；+；）将对象实例化为类属性 我用C++写了一个战舰的标准游戏，里面有一个包含两个玩家对象的游戏对象。当我尝试在游戏构造函数中实例化玩家对象时，IntelliSense给了我两个错误：

IntelliSense:表达式必须是可修改的左值

IntelliSense:不存在合适的构造函数将“Player（）”转换为“Player”

这是我的头文件：

class Player {
public:
    Player(string name);
    //More unrelated stuff (Get/Set methods and Attributes)
};


class Game {
public:
    Game(bool twoPlayer, string Player1Name, string Player2Name);
    //Get and Set methods (not included)
    //Attributes:
    Player Player1();
    Player Player2();
    int turn;
};

我对播放器构造函数的定义：

Player::Player(string name)
{
    SetName(name);
    //Initialize other variables that don't take input
{

以及给出错误的代码：

//Game constructor
Game::Game(bool twoPlayer, string Player1Name, string Player2Name)
{
    Player1 = Player(Player1Name);  //These two lines give the first error
    Player2 = Player(Player2Name);
    turn = 1;
}

//Game class Gets
int Game::GetTurn() { return turn; }
Player Game::GetPlayer1() { return Player1; }  //These two lines give the second error
Player Game::GetPlayer2() { return Player2; }

我做错了什么？我试过换衣服

Player1 = Player(Player1Name);
Player2 = Player(Player2Name);

到

---

还有很多其他的事情，但都不管用。提前多谢

问题似乎出在头文件中的类游戏中：

class Game {
public:
    Game(bool twoPlayer, string Player1Name, string Player2Name);
    //Get and Set methods (not included)
    //Attributes:
    Player Player1;// 
    Player Player2;//  
    int turn;
}; 

---

声明成员时请删除括号，否则您将创建名为Player1和Player2的函数，这些函数不带参数

问题似乎出在头文件中的类游戏中：

class Game {
public:
    Game(bool twoPlayer, string Player1Name, string Player2Name);
    //Get and Set methods (not included)
    //Attributes:
    Player Player1;// 
    Player Player2;//  
    int turn;
}; 

---

声明成员时请删除括号，否则您将创建名为Player1和Player2的函数，这些函数不带参数

Player1

和

Player2

都是函数。我假设，您希望它们成为一个成员变量

将游戏定义更改为：

class Game
{
public:
    Game(bool twoPlayer, string Player1Name, string Player2Name);

    //Get and Set methods (not included)

    //Attributes:
    Player Player1;
    Player Player2;
    int turn;
};

并使用初始化列表初始化您的成员：

Game::Game(bool twoPlayer, string Player1Name, string Player2Name)
: Player1(Player1Name)
, Player2(Player2Name)
, turn(1)
{
}

阅读更多有关为什么要初始化成员的信息：

Game::Game(bool twoPlayer, string Player1Name, string Player2Name)
: Player1(Player1Name)
, Player2(Player2Name)
, turn(1)
{
}

现在，这两条线：

Player Game::GetPlayer1() { return Player1; }
Player Game::GetPlayer2() { return Player2; }

---

将不再生成任何错误。

Player1

和

Player2

是函数。我假设，您希望它们成为一个成员变量

将游戏定义更改为：

class Game
{
public:
    Game(bool twoPlayer, string Player1Name, string Player2Name);

    //Get and Set methods (not included)

    //Attributes:
    Player Player1;
    Player Player2;
    int turn;
};

并使用初始化列表初始化您的成员：

Game::Game(bool twoPlayer, string Player1Name, string Player2Name)
: Player1(Player1Name)
, Player2(Player2Name)
, turn(1)
{
}

阅读更多有关为什么要初始化成员的信息：

Game::Game(bool twoPlayer, string Player1Name, string Player2Name)
: Player1(Player1Name)
, Player2(Player2Name)
, turn(1)
{
}

现在，这两条线：

Player Game::GetPlayer1() { return Player1; }
Player Game::GetPlayer2() { return Player2; }

---

将不再生成任何错误。

Player Player1（）是一个函数声明。
Player Player1（）是一个函数声明。啊！就这样！谢谢！啊！就这样！谢谢！