Question

我试图创建一个按钮，当我单击它时，它将游戏状态更改为正在玩，然后调用将要玩游戏的所有内容，但是我无法获得返回游戏状态的方法。

    private GameState Playbutton_Click(object sender, EventArgs e)
    {
        GameState CurrentState = GameState.playing;
        return CurrentState;
    }

我得到的错误是 “ Game1.GameState Game1.Playbutton_Click（object，EventArgs）”具有错误的返回类型

编辑：

在游戏1：

public class Game1 : Game
{
    GraphicsDeviceManager graphics;
    SpriteBatch spriteBatch;



    enum GameState
    {
        playing,
        menu,
        over
    }

    GameState CurrentState = GameState.playing;

    public Game1()
    {
        graphics = new GraphicsDeviceManager(this);
        Content.RootDirectory = "Content";
    }


    GameState CurrentState = GameState.playing;


              Playbutton = new Button(Content.Load<Texture2D>("button"), Content.Load<SpriteFont>("Font"))
        {
            position = new Vector2(690, 500),
            Text = ("Play")
        };
        Quitbutton = new Button(Content.Load<Texture2D>("button"), Content.Load<SpriteFont>("Font"))
        {
            position = new Vector2(690, 800),
            Text = ("Quit")
        };

        Playbutton.Click += Playbutton_Click;
        Quitbutton.Click += Quitbutton_Click;
        components = new List<Component>()
        {
            Playbutton,
            Quitbutton
        };


              private void Quitbutton_Click(object sender, EventArgs e)
    {
        Exit();
    }

    private void Playbutton_Click(object sender, EventArgs e)
    {
        GameState CurrentState = GameState.playing;
    }

按钮类：

   private MouseState _currentMouse;
    private SpriteFont _font;
    private bool _isHovering;
    private MouseState _previousMouse;
    private Texture2D _texture;

    public event EventHandler Click;
    public bool Clicked { get; private set; }
    public Color PenColour { get; set; }
    public Vector2 position { get; set; }
    public Rectangle Rectangle
    {
        get
        {
            return new Rectangle((int)position.X, (int)position.Y, _texture.Width, _texture.Height);
        }
    }
    public string Text { get; set; }
    public Button(Texture2D texture, SpriteFont font)
    {
        _texture = texture;
        _font = font;
        PenColour = Color.Black;
    }
    public override void Draw(GameTime gameTime, SpriteBatch spriteBatch)
    {
        var colour = Color.White;
        if (_isHovering)
            colour = Color.Gray;
        spriteBatch.Draw(_texture, Rectangle, colour);
        if (!string.IsNullOrEmpty(Text))
        {
            var x = (Rectangle.X + (Rectangle.Width / 2)) - (_font.MeasureString(Text).X / 2);
            var y = (Rectangle.Y + (Rectangle.Height / 2)) - (_font.MeasureString(Text).Y / 2);
            spriteBatch.DrawString(_font, Text, new Vector2(x, y), PenColour);
        }
    }
    public override void Update(GameTime gameTime)
    {
        _previousMouse = _currentMouse;
        _currentMouse = Mouse.GetState();
        var mouseRectangle = new Rectangle(_currentMouse.X, _currentMouse.Y, 1, 1);
        _isHovering = false;
        if (mouseRectangle.Intersects(Rectangle))
        {
            _isHovering = true;
            if (_currentMouse.LeftButton == ButtonState.Released &&      _previousMouse.LeftButton == ButtonState.Pressed)
              {
                Click?.Invoke(this, new EventArgs());
            }
        }
    }
}

Answer 1

事件处理程序的签名几乎总是：

void SomethingHappened(object sender, EventArgs e)

通过方法订阅此事件时，必须遵循其定义的方法签名，包括使用相同的返回类型，在本例中为void。

在这种情况下，您可以返回某些内容的唯一方法是创建一个变量，以后可以访问该变量并将其分配给它，或者（如果定义了事件处理程序，则）重新定义方法签名以返回{{1} }。

编辑：

根据您的链接代码，您似乎在修复中有正确的想法。按照惯例，您不应从EventHandler返回任何内容，而应拥有一个可以在触发事件时设置并稍后再访问它的成员变量：

GameState

返回类型错误的单人游戏

1 个答案:

编辑：