[Intel-gfx] Trying to understand the URB code

[Kenneth Graunke]

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