-1

我尝试使用SFML库编写代码,当我按下一个键时,我更改了一个 bool 变量。问题是我使用按键,我不知道如何实现按键释放,谁只工作 1 次 x 键,这就是我试图实现的原因,因为按键会影响多次改变变量。

if (Keyboard.IsKeyPressed(Keyboard.Key.A))
{
    values[0] = !values[0];
}
if (Keyboard.IsKeyPressed(Keyboard.Key.S))
{
    values[1] = !values[1];
}
if (Keyboard.IsKeyPressed(Keyboard.Key.D))
{
    values[2] = !values[2];
}

谢谢

4

1 回答 1

1

我认为您可以使用 Window.KeyPressed 和 Window.KeyReleased 并保存按下和释放的键

private Dictionary<Keyboard.Key, bool> keysArePressed = new Dictionary<Keyboard.Key, bool>
{
   {Keyboard.Key.A, false},
   {Keyboard.Key.S, false},
   {Keyboard.Key.D, false}
};

// Key
Window.KeyPressed += OnKeyPressed;
Window.KeyReleased += OnKeyReleased;

public void OnKeyPressed(object sender, KeyEventArgs e)
{
   if (e.Code == Keyboard.Key.A && !keysArePressed[Keyboard.Key.A])
   {
      Console.WriteLine("A pressed");
      keysArePressed[Keyboard.Key.A] = true;
   }
}

public void OnKeyReleased(object sender, KeyEventArgs e)
{
   if (e.Code == Keyboard.Key.A)
   {
      Console.WriteLine("A released");
      keysArePressed[Keyboard.Key.A] = false;
   }
}
于 2018-05-02T21:01:35.467 回答