🔒 Locked

HLSL Assembly Question

Started by yadango Aug 3, 2012 at 10:01 AM 4 replies 2.2k views

Original Post

Aug 3, 2012 at 10:01 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.
[source]mova a0.w, r4.x
mul r4.yzw, v0.y, c0[a0.w].xxyz
[/source]
is the equivalent expression
[source]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[/source]
or this?
[source]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[/source]
Thanks!

Digitalfragment

Aug 5, 2012 at 11:36 PM

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

yadango

Aug 5, 2012 at 11:57 PM

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

Erik Rufelt

Aug 6, 2012 at 12:41 AM

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.

yadango

Aug 6, 2012 at 1:24 AM

so maybe something like this?

that's why the mask has to be in order...

[source lang="cpp"]// mova a0.w, r4.x
temp.x = r4.x
temp.y = r4.x
temp.z = r4.x
temp.w = r4.x
a0.x = unchanged
a0.y = unchanged
a0.z = unchanged
a0.w = temp.w

// mul r4.yzw, v0.y, c0[a0.w].xxyz
temp1.x = v0.y
temp1.y = v0.y
temp1.z = v0.y
temp1.w = v0.y
temp2.x = c0[a0.w].x
temp2.y = c0[a0.w].x
temp2.z = c0[a0.w].y
temp2.w = c0[a0.w].z
r4.x = unchanged
r4.y = temp1.y * temp2.y
r4.z = temp1.z * temp2.z
r4.w = temp1.w * temp2.w[/source]

Erik Rufelt

Aug 6, 2012 at 9:41 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.

Topic Locked

This topic has been locked by a moderator. New replies are not allowed.