[Intel-gfx] [PATCH 12/13] drm/i915: Consolidate legacy semaphore initialization

[Dave Gordon]

On 15/07/16 14:13, Tvrtko Ursulin wrote:
>
> On 29/06/16 17:00, Chris Wilson wrote:
>> On Wed, Jun 29, 2016 at 04:41:58PM +0100, Tvrtko Ursulin wrote:
>>>
>>> On 29/06/16 16:34, Chris Wilson wrote:
>>>> On Wed, Jun 29, 2016 at 04:09:31PM +0100, Tvrtko Ursulin wrote:
>>>>> From: Tvrtko Ursulin <tvrtko.ursulin at intel.com>
>>>>>
>>>>> Replace per-engine initialization with a common half-programatic,
>>>>> half-data driven code for ease of maintenance and compactness.
>>>>>
>>>>> Signed-off-by: Tvrtko Ursulin <tvrtko.ursulin at intel.com>
>>>>
>>>> This is the biggest pill to swallow (since our 5x5 table is only
>>>> sparsely populated), but it looks correct, and more importantly
>>>> easier to
>>>> read.
>>>
>>> Yeah I was out of ideas on how to improve it. Fresh mind needed to
>>> try and spot a pattern in how MI_SEMAPHORE_SYNC_* and GEN6_*SYNC map
>>> to bits and registers respectively, and write it as a function.
>>
>> It's actually a very simple cyclic function based on register
>> offset = base + (signaler hw_id - waiter hw_id - 1) % num_rings.
>>
>> (The only real challenge is picking the direction.)
>>
>> commit c8c99b0f0dea1ced5d0e10cdb9143356cc16b484
>> Author: Ben Widawsky <ben at bwidawsk.net>
>> Date:   Wed Sep 14 20:32:47 2011 -0700
>>
>>      drm/i915: Dumb down the semaphore logic
>>
>>      While I think the previous code is correct, it was hard to follow
>> and
>>      hard to debug. Since we already have a ring abstraction, might as
>> well
>>      use it to handle the semaphore updates and compares.
>
> Doesn't seem to fit, or I just can't figure it out. Needs two functions
> to get rid of the table:
>
> f1(0, 1) = 2
> f1(0, 2) = 0
> f1(0, 3) = 2
> f1(1, 0) = 0
> f1(1, 2) = 2
> f1(1, 3) = 1
> f1(2, 0) = 2
> f1(2, 1) = 0
> f1(2, 3) = 0
> f1(3, 0) = 1
> f1(3, 1) = 1
> f1(3, 2) = 1
>
> and:
>
> f2(0, 1) = 1
> f2(0, 2) = 0
> f2(0, 3) = 1
> f2(1, 0) = 0
> f2(1, 2) = 1
> f2(1, 3) = 2
> f2(2, 0) = 1
> f2(2, 1) = 0
> f2(2, 3) = 0
> f2(3, 0) = 2
> f2(3, 1) = 2
> f2(3, 2) = 2
>
> A weekend math puzzle for someone? :)
>
> Regards,
> Tvrtko

Here's the APL expression for (the transpose of) f2, with -1's filled in 
along the leading diagonal (you need ⎕io←0 so the ⍳-vectors are in origin 0)

       {¯1+(⍵≠⍳4)⍀(2|⍵)⌽(⌽⍣(1=⍵))1+⍳3}¨⍳4

┌────────┬────────┬────────┬────────┐
│¯1 0 1 2│1 ¯1 0 2│0 1 ¯1 2│1 2 0 ¯1│
└────────┴────────┴────────┴────────┘

or transposed back so that the first argument is the row index and the 
second is the column index:

       ⍉↑{¯1+(⍵≠⍳4)⍀(2|⍵)⌽(⌽⍣(1=⍵))1+⍳3}¨⍳4

¯1  1  0  1
  0 ¯1  1  2
  1  0 ¯1  0
  2  2  2 ¯1

http://tryapl.org/?a=%u2349%u2191%7B%AF1+%28%u2375%u2260%u23734%29%u2340%282%7C%u2375%29%u233D%28%u233D%u2363%281%3D%u2375%29%291+%u23733%7D%A8%u23734&run

f1 is trivially derived from this by the observation that f1 is just f2 
with the 1's and 2's interchanged.

.Dave.