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[英]SelectedIndex returns -1 in void/static void function
[英]Is there any way to break out of a function that returns void?
我有一个函数,为给定另一个tile的特定tile设置绘制状态。 绘制状态将更改的图块比较周围的图块然后相应地更新。 我将尝试在下面说明
[b] [b] [a]
[b] [a] [a]
[a] [a] [a] where a = sand && b = water
当a检测到b正在接近它时,它必须更新其绘制状态。 所以我有一个适用于大写,小写,左大小写和右大小写的函数。 我现在需要修改该功能,以便它可以处理左右情况,右上角情况,右下角情况等。这是我的功能
public override void CompareBorderingTiles(Tile T)
{
if (T is Water)
{
float leftBound = location.X - (Tile.TileWidth * Tile.TileScale);
float rightBound = location.X + (Tile.TileWidth * Tile.TileScale);
float upperBound = location.Y - (Tile.TileHieght * Tile.TileScale);
float bottomBound = location.Y + (Tile.TileHieght * Tile.TileScale);
if (T.GridLocation.X == leftBound)
{
drawstate = DrawState.Left;
}
if (T.GridLocation.X == rightBound)
drawstate = DrawState.Right;
if (T.GridLocation.Y == upperBound)
drawstate = DrawState.Upper;
if (T.GridLocation.Y == bottomBound)
drawstate = DrawState.Lower;
}
base.CompareBorderingTiles(T);
}
对于为什么我想要突破这个功能,或者可能不是这个功能,它应该是非常明确的解释。 基本上我有一个枚举,告诉我我的绘制状态是什么(drawstate是枚举)。 谁能告诉我是否可以设置正确的绘制状态然后退出我的功能?
只需使用您想要结束的return语句:
return;
所以,在您的代码中,您可以这样做:
public override void CompareBorderingTiles(Tile T)
{
if (T is Water)
{
float leftBound = location.X - (Tile.TileWidth * Tile.TileScale);
float rightBound = location.X + (Tile.TileWidth * Tile.TileScale);
float upperBound = location.Y - (Tile.TileHieght * Tile.TileScale);
float bottomBound = location.Y + (Tile.TileHieght * Tile.TileScale);
if (T.GridLocation.X == leftBound)
{
drawstate = DrawState.Left;
return;
}
if (T.GridLocation.X == rightBound)
{
drawstate = DrawState.Right;
return;
}
if (T.GridLocation.Y == upperBound)
{
drawstate = DrawState.Upper;
return;
}
if (T.GridLocation.Y == bottomBound)
{
drawstate = DrawState.Lower;
return;
}
}
base.CompareBorderingTiles(T);
}
只需使用return;
就其本身而言,这将立即从函数“返回”。
虽然你真的需要回归吗?
if (T.GridLocation.X == leftBound)
{
drawstate = DrawState.Left;
}
else if (T.GridLocation.X == rightBound)
{
drawstate = DrawState.Right;
}
else if (T.GridLocation.Y == upperBound)
{
drawstate = DrawState.Upper;
}
else if (T.GridLocation.Y == bottomBound)
{
drawstate = DrawState.Lower;
}
这应该使代码在未来更容易维护。
您可以使用return
退出函数。
你可以使用return;
在函数的任何一点,它将在那里离开函数。 使用return意味着不会调用您的基函数。 如果您需要基本函数被调用使用else if
那么,当你条件满足,就不会检查剩余的if语句:
public override void CompareBorderingTiles(Tile T)
{
if (T is Water)
{
float leftBound = location.X - (Tile.TileWidth * Tile.TileScale);
float rightBound = location.X + (Tile.TileWidth * Tile.TileScale);
float upperBound = location.Y - (Tile.TileHieght * Tile.TileScale);
float bottomBound = location.Y + (Tile.TileHieght * Tile.TileScale);
if (T.GridLocation.X == leftBound)
{
drawstate = DrawState.Left;
}
else if (T.GridLocation.X == rightBound)
drawstate = DrawState.Right;
else if (T.GridLocation.Y == upperBound)
drawstate = DrawState.Upper;
else if (T.GridLocation.Y == bottomBound)
drawstate = DrawState.Lower;
}
base.CompareBorderingTiles(T);
}
使用return;
该函数将作为void返回
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