[Intel-gfx] Trying to understand the URB code
Kenneth Graunke
kenneth at whitecape.org
Thu Apr 7 07:01:06 CEST 2011
Hi Nanhai,
I'm trying to understand how the Gen6 URB setup works, and I had some
questions...
if (IS_GT1(intel->intelScreen->deviceID)) {
urb_size = 32 * 1024;
max_urb_entry = 128;
} else {
urb_size = 64 * 1024;
max_urb_entry = 256;
}
I see in vol5c.5 that GT1 has 32kB of URB space and GT2 has 64kB, so
urb_size must be the total size of the URB. But what is max_urb_entry?
Where do 128 and 256 come from?
brw->urb.vs_size = MAX2(brw->vs.prog_data->urb_entry_size, 1);
It looks like brw->vs.prog_data->urb_entry_size is the size of a single
VUE, which depends on the number of input/outputs in the particular
vertex shader being used.
So, brw->urb.vs_size is also the size of a VUE, but at least 1.
What are the units here? The number of 1024-bit blocks? (I'm looking
at 3DSTATE_URB in vol2a of the bspec...)
brw->urb.nr_vs_entries = max_urb_entry;
brw->urb.nr_gs_entries = max_urb_entry;
if (2 * brw->urb.vs_size * brw->urb.nr_vs_entries > urb_size)
brw->urb.nr_vs_entries = brw->urb.nr_gs_entries =
(urb_size ) / (2 * brw->urb.vs_size);
Here it looks like you're trying to allocate half of the URB to the VS,
and half to the GS. I'm confused by the units, though: if vs_size is in
1024-bit (128-byte) blocks and urb_size is in bytes, don't we need to
multiply vs_size by 128?
I think the above code could be simplified to:
int urb_entries = urb_size / (2 * brw->urb.vs_size * 128);
brw->urb.nr_vs_entries = brw->urb.nr_gs_entries = MIN2(urb_entries,
max_urb_entry);
What do you think?
Thanks for the help.
--Kenneth
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