[英]Metal RGB to YUV conversion compute shader
I am trying to write a Metal compute shader for converting from RGB to YUV, but am getting build errors.我正在尝试编写用于从 RGB 转换为 YUV 的 Metal 计算着色器,但遇到构建错误。
typedef struct {
float3x3 matrix;
float3 offset;
} ColorConversion;
// Compute kernel
kernel void kernelRGBtoYUV(texture2d<half, access::sample> inputTexture [[ texture(0) ]],
texture2d<half, access::write> textureY [[ texture(1) ]],
texture2d<half, access::write> textureCbCr [[ texture(2) ]],
constant ColorConversion &colorConv [[ buffer(0) ]],
uint2 gid [[thread_position_in_grid]])
{
// Make sure we don't read or write outside of the texture
if ((gid.x >= inputTexture.get_width()) || (gid.y >= inputTexture.get_height())) {
return;
}
float3 inputColor = float3(inputTexture.read(gid).rgb);
float3 yuv = colorConv.matrix*inputColor + colorConv.offset;
half2 uv = half2(yuv.gb);
textureY.write(half(yuv.x), gid);
if (gid.x % 2 == 0 && gid.y % 2 == 0) {
textureCbCr.write(uv, uint2(gid.x / 2, gid.y / 2));
}
}
The last line, ie write to textureCbCr throws an error:最后一行,即写入 textureCbCr 会抛出错误:
no matching member function for call to 'write'
According to the Metal Shading Language Specification, the first parameter of all overloads of write
on texture2d<>
are 4-element vectors.根据 Metal Shading Language Specification, texture2d<>
上所有write
重载的第一个参数是 4 元素向量。 This is the case even if the texture you're writing to has fewer than 4 components.即使您写入的纹理少于 4 个组件,情况也是如此。 So you can fix this by replacing the erroneous line with:因此,您可以通过将错误的行替换为以下内容来解决此问题:
textureCbCr.write(half4(yuv.xyzz), uint2(gid.x / 2, gid.y / 2));
And the superfluous components will be masked out when performing the write.并且在执行写入时将屏蔽掉多余的组件。
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