Can kernel based tile shader write to color attachment?

Question

Based on an example in Metal shading language spec:

Example:
struct Foo {
    float4 a [[color(0)]];
    int4 b [[color(1)]];
};

kernel void
my_kernel(imageblock<Foo, imageblock_layout_implicit> img_blk,
 ushort2 lid [[thread_index_in_threadgroup]] …) 
{
…
    Foo f = img_blk.read(lid); float4 r = f.a;
…
    f.a = r;
…
    img_blk.write(f, lid);
}

As the members of the imageblock alias color attachment I think after "imgblk.write(...)" the imageblock will be written to corresponding color attachment.

I experimented this on Apple's example: Forward-plus with tile shading, I tried to use imageblock.write(..) to write to the color attachments in the tile shader but I got very weird results:

only pixels of the background are changed, but the result is too much darker than what I set, e.g. I set color=float4(1,0,0,1) but on screen it's float4(0.057, 0, 0, 1)
color of other parts strangely depends on whether/what is written to the imageblock in the previous fragment shader pass, consider I set the values in the imageblock to constants.

Anyway it feels like imageblock.write() doesn't work right in tile shader. Or how to use it correctly?

painkiller · Accepted Answer · 2020-06-03 03:33:54Z

0

Solved, there are several things lead to the weird result I got: 1. The color format was srgb 2. The numSamples the example project used were 4 3. I did imageblock write without specifying imageblock_data_rate

In the case of multisampling, one needs to read/write the imageblock for each sample

answered Jun 3, 2020 at 3:33

painkiller

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leonheess Over a year ago

If you answered your own question, please mark it as such so the question is removed from the unanswered pile. Thank you

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