本文整理匯總了C#中OpenRA.Graphics.WorldRenderer.ProjectedPosition方法的典型用法代碼示例。如果您正苦於以下問題：C# WorldRenderer.ProjectedPosition方法的具體用法？C# WorldRenderer.ProjectedPosition怎麽用？C# WorldRenderer.ProjectedPosition使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類OpenRA.Graphics.WorldRenderer的用法示例。
在下文中一共展示了WorldRenderer.ProjectedPosition方法的2個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的C#代碼示例。
示例1: Render
public IEnumerable<IRenderable> Render(Actor self, WorldRenderer wr)
{
var pos = wr.ScreenPxPosition(self.CenterPosition);
var bounds = self.Bounds;
bounds.Offset(pos.X, pos.Y);
var spaceBuffer = (int)(10 / wr.Viewport.Zoom);
var effectPos = wr.ProjectedPosition(new int2(pos.X, bounds.Y - spaceBuffer));
yield return new TextRenderable(font, effectPos, 0, color, name);
}示例2: DrawZap
static IEnumerable<IFinalizedRenderable> DrawZap(WorldRenderer wr, float2 from, float2 to, ISpriteSequence s, out float2 p, string palette)
{
var dist = to - from;
var q = new float2(-dist.Y, dist.X);
var c = -float2.Dot(from, q);
var rs = new List<IFinalizedRenderable>();
var z = from;
var pal = wr.Palette(palette);
while ((to - z).X > 5 || (to - z).X < -5 || (to - z).Y > 5 || (to - z).Y < -5)
{
var step = steps.Where(t => (to - (z + new float2(t[0], t[1]))).LengthSquared < (to - z).LengthSquared)
.MinBy(t => Math.Abs(float2.Dot(z + new float2(t[0], t[1]), q) + c));
var pos = wr.ProjectedPosition((z + new float2(step[2], step[3])).ToInt2());
rs.Add(new SpriteRenderable(s.GetSprite(step[4]), pos, WVec.Zero, 0, pal, 1f, true).PrepareRender(wr));
z += new float2(step[0], step[1]);
if (rs.Count >= 1000)
break;
}
p = z;
return rs;
}