On 20/07/16 13:50, Dave Gordon wrote:
On 20/07/16 10:54, Tvrtko Ursulin wrote:On 19/07/16 19:38, Dave Gordon wrote: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.ursu...@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.ursu...@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 <b...@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, TvrtkoHere'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:-C ! How to convert that to C ? :)f1 is trivially derived from this by the observation that f1 is just f2 with the 1's and 2's interchanged.Ah yes, nicely spotted. Regards, TvrtkoAssuming you don't care about the leading diagonal (x == y), then (⍵≠⍳4)⍀(2|⍵)⌽(⌽⍣(1=⍵)) translates into: int f2(unsigned int x, unsigned int y) { x -= x >= y; if (y == 1) x = 3 - x; x += y & 1; return x % 3; } y:x 0 1 2 3 0: 0 0 1 2 1: 1 1 0 2 2: 0 1 1 2 3: 1 2 0 0 Each line of C corresponds quite closely to one operation in the APL :) Although, in APL we tend to leave the data unchanged while shuffling it around into new shapes, whereas the C below does the equivalent things by changing the data (noting that it's all modulo-3 arithmetic). (⍵≠⍳4)⍀ inserts the leading diagonal, corresponding to the subtraction of x >= y (which removes the leading diagonal). ⌽⍣(1=⍵) reverses the sequence if y==1; in C, that's the 3-x (2|⍵)⌽ rotates the sequence by 1 if y is odd; that's the += and the final % ensures that the result is 0-2.
I was hoping for a solution which does not include conditionals, someone led me to believe it is possible! :)
But thanks, your transformation really works. I've sent a patch implementing it to trybot for now.
Regards, Tvrtko _______________________________________________ Intel-gfx mailing list Intel-gfx@lists.freedesktop.org https://lists.freedesktop.org/mailman/listinfo/intel-gfx
