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++ 编写一个标准的战舰游戏，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 给出两个错误：

IntelliSense：表达式必须是可修改的左值

IntelliSense：不存在合适的构造函数可从“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; }

我究竟做错了什么？我尝试过改变

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

到

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

以及其他一些事情，但没有任何作用。预先非常感谢！

Answer 1

Player1和Player2是函数。我假设您希望它们成为成员变量。

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将定义更改Game为：

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

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并使用初始化列表来初始化您的成员：

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

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详细了解为什么应该初始化成员：

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• 构造函数初始化与赋值
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• 我的构造函数应该使用 \xe2\x80\x9c 初始化列表 \xe2\x80\x9d 或 \xe2\x80\x9c 赋值 \xe2\x80\x9d 吗？
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现在，这两行：

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

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不会再产生任何错误。

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