[Mesa-dev] [PATCH] glsl: Skip invariant/precision linker checks for built-in variables.

[Ian Romanick]

On 10/19/2016 02:31 PM, Kenneth Graunke wrote:
> On Wednesday, October 19, 2016 1:40:39 PM PDT Ian Romanick wrote:
>> On 10/19/2016 11:11 AM, Kenneth Graunke wrote:
>>> Brian found a bug with my "inline built-ins immediately" code for shaders
>>> which use ftransform() and declare gl_Position invariant:
>>>
>>> https://lists.freedesktop.org/archives/mesa-dev/2016-October/132452.html
>>>
>>> Before my patch, things worked due to a specific order of operations:
>>>
>>> 1. link_intrastage_varyings imported the ftransform function into the VS
>>> 2. cross_validate_uniforms() ran and signed off that everything matched
>>> 3. do_common_optimization did both inlining and invariance propagation,
>>>    making the VS/FS versions of gl_ModelViewProjectionMatrix have
>>>    different invariant qualifiers...but after the check in step 2,
>>>    so we never raised an error.
>>>
>>> After my patch, ftransform() is inlined right away, and at compile time,
>>> do_common_optimization propagates the invariant qualifier to the
>>> gl_ModelViewProjectionMatrix.  When the linker eventually happens, it
>>> detects the mismatch.
>>
>> Why are we marking a uniform as invariant in the first place?  That
>> sounds boats.
> 
> I agree.  Propagating invariant/precise to ir_variables used in
> expressions is pretty bogus.  We should really track this with an
> ir_expression::exact flag similar to NIR's "exact" flag on ALU
> expressions.  I don't recall why it was done this way.
> 
> I've hit problems with invariant on uniforms before.  I tried not
> propagating it to uniforms back in the day and Curro convinced me
> it was wrong and would break things.
> 
> This is a hack - it fixes some cases, but won't fix them all.
> Not propagating to uniforms will fix some cases, but I think it
> breaks others.  I don't think we have tests for those cases.

Right... I think the problem case would be expression trees that involve
only uniforms wouldn't necessarily get the invariant treatment.  If one
shader had

    uniform mat4 u0;
    uniform mat4 u1;
    invariant out vec4 o;
    in vec4 i;

    void main()
    {
        o = u0 * u1 * i;
    }

and the other shader had

    uniform mat4 u0;
    uniform mat4 u1;
    invariant out vec4 o0;
    out vec4 o1;
    in vec4 i;

    void main()
    {
        o0 = u0 * u1 * i;
        o1 = u0 * i;
    }

The 'u0 * u1' part of the invariant expression might get optimized
differently in each shader.

Let's go with Ken's patch for now.

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