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HLSL Assembly Question


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#1 yadango   Members   -  Reputation: 559

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Posted 03 August 2012 - 04:01 AM

Hello,
I'm trying to learn HLSL assembly, using MSDN as a guide and a book or two. Most swizzle examples given are kind of simple and I saw some code that looked like this and wondered what the equivalent statements were.
mova a0.w, r4.x

mul r4.yzw, v0.y, c0[a0.w].xxyz


is the equivalent expression
a0.w = r4.x

r4.y = v0.y * c0[a0.w].x

r4.z = v0.y * c0[a0.w].x

r4.w = v0.y * c0[a0.w].y

or this?
a0.w = r4.x

r4.y = v0.y * c0[a0.w].x

r4.z = v0.y * c0[a0.w].x

r4.w = v0.y * c0[a0.w].y

r4.w = v0.y * c0[a0.w].z

Thanks!

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#2 Digitalfragment   Members   -  Reputation: 747

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Posted 05 August 2012 - 05:36 PM

The first. the z in c0[a0.w].xxyz would get dropped as it doesn't have a corresponding output in r4.yzw

#3 yadango   Members   -  Reputation: 559

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Posted 05 August 2012 - 05:57 PM

Cool, thanks, makes sense now! The repeat rule only applies to the source register and not the dest.

#4 Erik Rufelt   Crossbones+   -  Reputation: 3031

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Posted 05 August 2012 - 06:41 PM

I'm not sure.. but I would guess neither.. and that r4.yzw = v0.y * c0[a0.w].xyz
The docs http://msdn.microsoft.com/en-us/library/windows/desktop/hh447193(v=vs.85).aspx says the destination register has a writemask.. which is not the same as a swizzle. I interpret it as that the calculation is done on 4 components and the components of the result is written to the destination register or skipped depending on the mask..
That is, xxyz * v0.y is calculated to a temporary result xyzw, and the x is skipped on write to destination since it's not in the mask.

#5 yadango   Members   -  Reputation: 559

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Posted 05 August 2012 - 07:24 PM

so maybe something like this? Posted Image that's why the mask has to be in order...

[source lang="cpp"]// mova a0.w, r4.xtemp.x = r4.xtemp.y = r4.xtemp.z = r4.xtemp.w = r4.xa0.x = unchangeda0.y = unchangeda0.z = unchangeda0.w = temp.w// mul r4.yzw, v0.y, c0[a0.w].xxyztemp1.x = v0.ytemp1.y = v0.ytemp1.z = v0.ytemp1.w = v0.ytemp2.x = c0[a0.w].xtemp2.y = c0[a0.w].xtemp2.z = c0[a0.w].ytemp2.w = c0[a0.w].zr4.x = unchangedr4.y = temp1.y * temp2.yr4.z = temp1.z * temp2.zr4.w = temp1.w * temp2.w[/source]

Edited by yadango, 05 August 2012 - 07:29 PM.


#6 Erik Rufelt   Crossbones+   -  Reputation: 3031

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Posted 06 August 2012 - 03:41 AM

The internal steps will differ from platform to platform, HLSL ASM is an intermediate format, but the end result is most likely like that.




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