状态改变法

Question

这只是一个简单的菜鸟问题... 我经常使用类似这样的东西（代码）来更改我的 GUI，所以我的问题是是否有比使用 bool 变量更有用的东西？

谢谢！

//Unity3D - C#

public class GuiBehaviour : MonoBehaviour
{
  private bool lookInside = false;

  void OnGUI ()
  {
    if (!lookInside) {
        if (GUILayout.Button ("Look Inside")) {
            lookInside = true;
        }
    } else {
        if (GUILayout.Button ("Exit View")) {
            lookInside = false;
        }
      }
  }
}

Answer 1

怎么样：

lookInside =!lookInside;

if(lookInside)
{
  GUILayout.Button ("Look Inside")
}
else
{
   GUILayout.Button ("Exit View")
}

Answer 2

对不同的按钮使用不同的处理程序，假设按钮仅在任一视图中可见。

private void InitializeComponent()
{
        lookInsideButton = new System.Windows.Forms.Button();
        lookInsideButton.Click += new EventHandler(lookInsideButton_Click);

        exitViewButton = new System.Windows.Forms.Button();
        exitViewButton.Click += new EventHandler(exitViewButton_Click);
}

void lookInsideButton_Click(object sender, EventArgs e)
{
    ShowInsideView();
}

void exitViewButton_Click(object sender, EventArgs e)
{
    ExitInsideView();
}

System.Windows.Forms.Button lookInsideButton, exitViewButton;

Answer 3

我使用枚举是为了更好的可读性并拥有两个以上的状态。例如：

public class GuiBehaviour : MonoBehaviour
{
  private GUIState CurrentState;
  enum GUIState
  {
    LookInside,
    ExitView,
    GameOverView
  }

  void OnGUI(GUIState state)
  {
    CurrentState = state;
    switch(state)
    {
       case GUIState.LookInside:
         GUILayout.Button("Look Inside");
         break;
       case GUIState.ExitView:
         GUILayout.Button("Exit View");
         break;
       case GUIState.LookInside:
         GUILayout.Button("Game over");
         break;
     }
  }
}