问题
For example, if I want to do a grayscale transformation, I need to set up my threadsPerGroup and thread group in the following way.
NSUInteger maxTotalThreadsPerThreadgroup = [self.computePipelineState maxTotalThreadsPerThreadgroup];
MTLSize threadgroupCounts = MTLSizeMake(threadExecutionWidth * 2, threadExecutionWidth * 2, 1);
MTLSize threadsPerThreadGroup = MTLSizeMake([self.texutre width] / threadgroupCounts.width + 1,
[self.texutre height] / threadgroupCounts.height + 1,
1);
I know the image will be chopped into different blocks and each one will be processed by one thread group. But it seems in the kernel, we will just read the 2d texture, and then output the processed texture.
But the question is that how the image is chopped into different 2d texture? How do we know if each block of image get assigned to a thread to process? Is this done by Metal itself? Or we need to manually assign each block to each threadgroup by using the gid?
回答1:
Metal doesn't know or care whether your shader is operating on an image. It doesn't "chop" the image or anything like that.
A compute shader is processed over a "grid". The grid is an abstraction. It's an arbitrary way for you to organize the work. Metal doesn't assign any significance to the grid, such as associating a position in the grid with a pixel in an image.
Such an association, if it exists, is implicit in how your shader code behaves. Yes, that is largely based on what the shader does with thread_position_in_grid, thread_position_in_threadgroup, thread_index_in_threadgroup, etc.
So, if you're using a gid variable with the thread_position_in_grid attribute, and you use its coordinates as image coordinates, then that usage is what dictates that each grid position corresponds to an image pixel. Once you do that, then it follows that each thread group corresponds to a block of the image, since a thread group is just a block of grid positions. Again, though, this is not something that Metal is doing, it's something that your shader is doing.
You could do something entirely different and Metal wouldn't care.
来源：https://stackoverflow.com/questions/55768144/how-metal-distribute-the-image-block-to-each-thread-group