android - Canvas.clipPath(Path) 未按预期剪切

Question

我正在尝试将 Canvas 绘图操作剪辑为弧形楔形。但是，在将剪切路径设置为 Canvas 后，我没有得到预期的结果。

为了说明，这是我正在做的事情:

path.reset();

//Move to point #1
path.moveTo(rect.centerX(), rect.centerY());

//Per the documentation, this will draw a connecting line from the current
//position to the starting position of the arc (at 0 degrees), add the arc
//and my current position now lies at #2.
path.arcTo(rect, 0, -30);

//This should then close the path, finishing back at the center point (#3)
path.close();

这行得通，当我简单地绘制这条路径(canvas.drawPath(path,paint))时，它会绘制如上所示的楔形。但是，当我将此路径设置为 Canvas 的剪切路径并绘制到其中时:

//I've tried it with and without the Region.Op parameter
canvas.clipPath(path, Region.Op.REPLACE);
canvas.drawColor(Color.BLUE);

我得到以下结果(留下楔形只是为了显示引用):

因此，它似乎被剪辑到 Path 的边界矩形，而不是 Path 本身。有什么想法吗？

编辑 作为更新，我发现了一种更有效的方法，它允许硬件加速。首先，将整个图像(您将要剪裁的)绘制到屏幕外位图中。使用此 Bitmap 制作一个 BitmapShader，将该着色器设置为 Paint，然后使用该 Paint 绘制楔形路径对象:

drawMyBitmap(bitmap);
Shader shader = new BitmapShader(bitmap, Shader.TileMode.CLAMP, Shader.TileMode.CLAMP);
Paint paint = new Paint(Paint.ANTI_ALIAS_FLAG);
paint.setShader(shader);

@Override
public void onDraw(Canvas canvas) {
    canvas.drawArc(rect,         //The rectangle bounding the circle
                   startAngle,   //The angle (CW from 3 o'clock) to start
                   sweepAngle,   //The angle (CW from 3 o'clock) of the arc
                   true,         //Boolean of whether to draw a filled arc (wedge)
                   paint         //The paint with the shader attached
    );
}

Answer 1

您是使用 HC 或更高版本还是使用硬件加速？

如果是这样，clipPath 不受支持且存在问题。

developer.android.com/guide/topics/graphics/hardware-accel.html .