HLSL legacy samplers cause invalid output.

[slembcke]

If a HLSL shader declares a sampler2D uniform, that uniform is dropped when cross-compiling to GLSL using glslangValidator/spirv-cross.

For example, this is my input HLSL:

struct FragInput {
	float4 position : SV_POSITION;
	float2 uv : TEXCOORD0;
};

uniform sampler2D tex;

float4 FShader(in FragInput IN) : SV_TARGET0 {
	return tex2D(tex, IN.uv);
}

This is the resulting GLSL:

#version 450

layout(location = 0) in vec2 IN_uv;
layout(location = 0) out vec4 _entryPointOutput;

void main()
{
    _entryPointOutput = texture(tex, IN_uv.x);
}

Since I can run GLSL shaders that use DX9 style samplers through spirv-opt and back, I figured it was a glslangValidator bug. I compiled both the HLSL and the expected GLSL output and disassembled them. The only functional difference I could find was that the HLSL version had this line:

        %20 = OpTypeSampler

While the GLSL version had these two lines:

         %10 = OpTypeImage %float 2D 0 0 0 1 Unknown
         %20 = OpTypeSampledImage %10

Both versions perform the texture lookup using OpImageSampleImplicitLod, but according to the SPIR-V docs for it, only the OpTypeSampledImage argument is correct to pass to it. spirv-val seems to agree:

error: 34: Expected Sampled Image to be of type OpTypeSampledImage: ImageSampleImplicitLod

If I swap it out for those two lines, reassemble and cross compile it to GLSL, then the output correctly contains the sampler declaration and passes validation.

[slembcke]

Hmm. Seems like this isn't implemented and was supposed to throw an error or something:
https://github.com/KhronosGroup/glslang/blob/master/hlsl/hlslGrammar.cpp?utf8=%E2%9C%93#L274

[johnkslang]

I think there are some conflating comments here, as the DX9 path to the unimplemented(...) might not be taken (and some other things). But, fundamentally, yes, glslang is not supporting DX9.

This is captured in #362. So, if that's the real issue here, we can close this.