On 12.01.2015 01:57, Ilia Mirkin wrote:
On Sun, Jan 11, 2015 at 7:43 PM, Tobias Klausmann
wrote:
On 11.01.2015 06:05, Ilia Mirkin wrote:
Can you elaborate a bit as to why that's the right thing to do?
On Wed, Jan 7, 2015 at 1:52 PM, Tobias Klausmann
wrote:
If we capture transform feedback from n stream in (n-1) buffers we face a
NULL buffer, use the buffer (n-1) to capture the output of stream n.
This fixes one piglit test with nvc0:
arb_gpu_shader5-xfb-streams-without-invocations
Signed-off-by: Tobias Klausmann
---
src/mesa/state_tracker/st_cb_xformfb.c | 5 +
1 file changed, 5 insertions(+)
diff --git a/src/mesa/state_tracker/st_cb_xformfb.c
b/src/mesa/state_tracker/st_cb_xformfb.c
index 8f75eda..5a12da4 100644
--- a/src/mesa/state_tracker/st_cb_xformfb.c
+++ b/src/mesa/state_tracker/st_cb_xformfb.c
@@ -123,6 +123,11 @@ st_begin_transform_feedback(struct gl_context *ctx,
GLenum mode,
struct st_buffer_object *bo =
st_buffer_object(sobj->base.Buffers[i]);
if (bo) {
+ if (!bo->buffer)
+/* If we capture transform feedback from n streams into
(n-1)
+ * buffers we have to write to buffer (n-1) for stream n.
+ */
+bo = st_buffer_object(sobj->base.Buffers[i-1]);
/* Check whether we need to recreate the target. */
if (!sobj->targets[i] ||
sobj->targets[i] == sobj->draw_count ||
--
2.2.1
Quoted from Ilia Mirkin, to specify what shall be elaborated:
"Can you explain (on-list) why using buffer n - 1 is the right thing to
do to capture output of stream n? I would have thought that the output
for that stream should be discarded or something.
Like with a spec quotation or some other justification. i.e. why is
the code you wrote correct? Why is it better than, say, bo =
buffers[0], or some other thing entirely?"
Yeah thats the most concerning point i see as well. The problem is that
there is a interaction between arb_gpu_shader5 and arb_transform_feedback3,
but after a bit of reading i think the patch is actually what we should do:
From the arb_transfrom_feedback3 spec:
"
(3) How might you use transform feedback with geometry shaders and
multiple vertex streams?
RESOLVED: As a simple example, let's say you are processing triangles
and capture both processed triangle vertices and some values that are
computed per-primitive (e.g., facet normal). The geometry shader
might declare its outputs like the following:
layout(stream = 0) out vec4 position;
layout(stream = 0) out vec4 texcoord;
layout(stream = 1) out vec4 normal;
"position" and "texcoord" would be per-vertex attributes written to
vertex stream 0; "normal" would be a per-triangle facet normal. The
geometry shader would emit three vertices to stream zero (the
processed
input vertices) and a single vertex to stream one (the per-triangle
data). The transform feedback API usage for this case would be
something like:
// Set up buffer objects 21 and 22 to capture data for per-vertex
and
// per primitive values.
glBindBufferBase(GL_TRANSFORM_FEEDBACK_BUFFER, 0, 21);
glBindBufferBase(GL_TRANSFORM_FEEDBACK_BUFFER, 1, 22);
// Set up XFB to capture position and texcoord to buffer binding
// point 0 (buffer 21 bound), and normal to binding point 1 (buffer
// 22 bound).
char *strings[] = { "position", "texcoord", "gl_NextBuffer",
"normal" };
"
-> Especially the comments are enlightening as to where the outputs should
go. Thats what happens with the
"arb_gpu_shader5-xfb-streams-without-invocations" test, where two
stream(outputs) are captured into one buffer.
One might argue now if we have to count .Buffers[i-1] for all buffers after
this...
Comments and additional feedback is always appreciated!
The thing you're quoting is talking about the case where everything's
supposed to work. I haven't investigated, but I'm guessing that the
test has a layout(stream=1) but no buffer is bound at index 1.
Actually no, the layout reads like this:
layout(stream = 0) out float stream0_0_out;
layout(stream = 1) out vec2 stream1_0_out;
layout(stream = 2) out float stream2_0_out;
layout(stream = 2) out vec4 stream2_1_out;
where no buffer is bound to stream2_1_out.
Is that right? In that case, I would imagine that the TF output should
actually just get dropped on the floor. I would assume that this is in
the ARB_tf3 spec, but I don't have time to go digging right now.
As the layout read more like the one from the specs, i'd still go with
[n-1] and honestly (not proven correct) i assume that a stream without
an output is just wrong and should not even get through the linker...
Tobias
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