使用自定义构造函数将对象实例化为类属性 (C++)

Question

我正在使用 C++ 编写一个标准的 Battleship 游戏，其中包含一个 Game 对象，其中包含两个 Player 对象。当我尝试在 Game 构造函数中实例化 Player 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来将“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 Player 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？我试过改变

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

到

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

还有很多其他的东西，但没有任何效果。提前非常感谢！

Answer 1

Player1 和 Player2 是函数。我假设，您希望它们成为成员变量。

将Game 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;
};

并使用初始化列表来初始化您的成员：

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

阅读更多关于为什么你应该初始化你的成员：

• Constructor initialization vs assignment
• Should my constructors use “initialization lists” or “assignment”?

现在，这两行：

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

不会再产生任何错误。

Answer 2

问题似乎出在头文件中的 Game 类中：

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

在声明成员时删除括号，否则您将创建名为 Player1 和 Player2 且不带参数的函数