Bug 24781

Summary: Incorrect calculation of free dwords in ring buffer
Product: DRI Reporter: Rafał Miłecki <zajec5>
Component: DRM/RadeonAssignee: Default DRI bug account <dri-devel>
Status: RESOLVED NOTABUG QA Contact:
Severity: normal    
Priority: medium    
Version: XOrg git   
Hardware: Other   
OS: All   
Whiteboard:
i915 platform: i915 features:
Attachments:
Description Flags
drm/radeon/kms: fix calculation of free dwords in ring none

Description Rafał Miłecki 2009-10-28 17:53:56 UTC 
Created attachment 30782 [details] [review]
drm/radeon/kms: fix calculation of free dwords in ring

There are two scenarios of read/write pointers relation:

1) read lower than write
/----------------------------\
|     A B C D E F G          |
| 1 2 3 4 5 6 7 8 9 10 11 12 |
\----------------------------/
      R           W
used = w - r
free = size - used
free = size - (w - r)
free = size - w + r
free = r + size - w


2) write lower than read
/----------------------------\
| E F G           A  B  C  D |
| 1 2 3 4 5 6 7 8 9 10 11 12 |
\----------------------------/
      W           R
free = r - w


I made there ascii pictures to make it clear, as noone believed me today on IRC we have some bug there ;)

Problem is we calculate free dwords for first situation only (always). We have to detect second situation and use other calculation then.
Comment 1 Rafał Miłecki 2009-10-28 17:57:52 UTC 
You can see real hitting this issue in bug #24535 (check rings dumps logs).
Comment 2 Alex Deucher 2009-10-28 23:27:00 UTC 
The current code works as is since the ring size is always a power of two as per the comment in the code:

      A B C D E F G          
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
      R           W

(3 + 16 - 9) & 15 = 10

E F G                      A  B  C  D 
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
    W                      R

(12 + 16 - 2) & 15 = 10
Comment 3 Rafał Miłecki 2009-10-28 23:31:25 UTC 
(In reply to comment #2)
> The current code works as is since the ring size is always a power of two as
> per the comment in the code:

It seems so... I've never played with so hackish (like for me) bit operations. Thanks for explaining.

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