Libgdx的弧函数不是绘制弧形,而是绘制一个饼图(即有两条线连接到弧的起点)。
shapeRenderer.begin(ShapeType.Line);
shapeRenderer.arc(x, y, radius, 30, 120);
shapeRenderer.end();
有没有解决方案可以让libgdx绘制类似于html5画布弧线函数的弧曲线?
/** Draws an arc using {@link ShapeType#Line} or {@link ShapeType#Filled}. */
public void arc (float x, float y, float radius, float start, float degrees, int segments) {
if (segments <= 0) throw new IllegalArgumentException("segments must be > 0.");
float colorBits = color.toFloatBits();
float theta = (2 * MathUtils.PI * (degrees / 360.0f)) / segments;
float cos = MathUtils.cos(theta);
float sin = MathUtils.sin(theta);
float cx = radius * MathUtils.cos(start * MathUtils.degreesToRadians);
float cy = radius * MathUtils.sin(start * MathUtils.degreesToRadians);
if (shapeType == ShapeType.Line) {
check(ShapeType.Line, ShapeType.Filled, segments * 2 + 2);
renderer.color(colorBits);
renderer.vertex(x, y, 0); <b><--- CENTER</b>
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0); <b><--- LINE TO START POINT</b>
for (int i = 0; i < segments; i++) {
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0);
float temp = cx;
cx = cos * cx - sin * cy;
cy = sin * temp + cos * cy;
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0);
}
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0); <b><-- LINE TO END POINT</b>
...
CENTER
和 LINE TO END POINT
,以及每行前面的 renderer.color(..
代码。LINE TO START POINT
- 它似乎设置曲线的起点,但实际上在循环内部也完成了这一点,因此是多余的。)public class Arc extends ShapeRenderer{
private final ImmediateModeRenderer renderer;
private final Color color = new Color(1, 1, 1, 1);
public Arc(){
renderer = super.getRenderer();
}
/** Draws an arc using {@link ShapeType#Line} or {@link ShapeType#Filled}. */
public void arc (float x, float y, float radius, float start, float degrees) {
int segments = (int)(6 * (float)Math.cbrt(radius) * (degrees / 360.0f));
if (segments <= 0) throw new IllegalArgumentException("segments must be > 0.");
float colorBits = color.toFloatBits();
float theta = (2 * MathUtils.PI * (degrees / 360.0f)) / segments;
float cos = MathUtils.cos(theta);
float sin = MathUtils.sin(theta);
float cx = radius * MathUtils.cos(start * MathUtils.degreesToRadians);
float cy = radius * MathUtils.sin(start * MathUtils.degreesToRadians);
for (int i = 0; i < segments; i++) {
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0);
float temp = cx;
cx = cos * cx - sin * cy;
cy = sin * temp + cos * cy;
renderer.color(colorBits);
renderer.vertex(x + cx, y + cy, 0);
}
}
}
然后这样调用它
Arc a = new Arc();
a.setProjectionMatrix(cam.combined);
a.begin(ShapeType.Line);
a.arc(10, 10, 10, 30, 120);
a.end();
arc.setColor()
指令。color
的值应该通过调用 setColor()
进行更新。不过做得很好。我已经向您的源代码中添加了一个可工作的实现。 - EntangledLoops/** Draws an arc with 'stroke' of given width */
fun ShapeRenderer.strokeArc(strokeWidth: Float, x: Float, y: Float, radius: Float, start: Float, degrees: Float, sampling: Float = 2f, color: Color = Color.WHITE) {
val segments = ((6 * Math.cbrt(radius.toDouble()) * (Math.abs(degrees) / 360.0f)) * sampling).toInt()
val colorBits = color.toFloatBits()
for (i in 0 until segments) {
val x1 = radius * MathUtils.cosDeg(start + (degrees / segments) * i)
val y1 = radius * MathUtils.sinDeg(start + (degrees / segments) * i)
val x2 = (radius - strokeWidth) * MathUtils.cosDeg(start + (degrees / segments) * i)
val y2 = (radius - strokeWidth) * MathUtils.sinDeg(start + (degrees / segments) * i)
val x3 = radius * MathUtils.cosDeg(start + (degrees / segments) * (i + 1))
val y3 = radius * MathUtils.sinDeg(start + (degrees / segments) * (i + 1))
val x4 = (radius - strokeWidth) * MathUtils.cosDeg(start + (degrees / segments) * (i + 1))
val y4 = (radius - strokeWidth) * MathUtils.sinDeg(start + (degrees / segments) * (i + 1))
renderer.color(colorBits)
renderer.vertex(x + x1, y + y1, 0f)
renderer.color(colorBits)
renderer.vertex(x + x3, y + y3, 0f)
renderer.color(colorBits)
renderer.vertex(x + x2, y + y2, 0f)
renderer.color(colorBits)
renderer.vertex(x + x3, y + y3, 0f)
renderer.color(colorBits)
renderer.vertex(x + x2, y + y2, 0f)
renderer.color(colorBits)
renderer.vertex(x + x4, y + y4, 0f)
}
}
我认为你和其他回答者已经发布了当前最佳解决方案。对于任何人的另一种“hack”:
可以在绘制的两条线上方渲染相同颜色的背景线,以便在之后覆盖它们。是的,这很差劲,但是是快速而简单的小型代码解决方案。