事实证明这是一个平等的更快的方法。

It turns out there's an even faster approach.

使用@ elmiguelao答案中的建议，我修改了片段着色器来进行像素交换。这允许我从saveFrame（）中删除交换代码。由于我不再需要内存中像素的临时副本，因此我完全取消了 int [] 缓冲区，从此切换：

Using the suggestion in @elmiguelao's answer, I modified the fragment shader to do the pixel swap. This allowed me to remove the swap code from saveFrame(). Since I no longer needed a temporary copy of the pixels in memory, I eliminated the int[] buffer entirely, switching from this:

int[] colors = [... copy from mPixelBuf, swap ...]
Bitmap.createBitmap(colors, mWidth, mHeight, Bitmap.Config.ARGB_8888);

到此：

Bitmap bmp = Bitmap.createBitmap(mWidth, mHeight, Bitmap.Config.ARGB_8888);
bmp.copyPixelsFromBuffer(mPixelBuf);

一旦我这样做，我的所有颜色都错了。

As soon as I did that, all of my colors were wrong.

事实证明 Bitmap＃copyPixelsFromBuffer（）想要RGBA顺序的像素，不 ARGB顺序。来自 glReadPixels（）的值已经采用正确的格式。因此，通过这种方式，我避免交换，避免不必要的副本，并且根本不需要调整片段着色器。

It turns out that Bitmap#copyPixelsFromBuffer() wants the pixels in RGBA order, not ARGB order. The values coming out of glReadPixels() are already in the right format. So by doing it this way I avoid the swap, avoid an unnecessary copy, and don't need to tweak the fragment shader at all.