uxa: Don't treat prepare_access as a flush synchronisation point.

The kernel will only emit a flush iff the buffer is currently owned by
the GPU. Instead of presuming that the kernel must emit a flush, it is
safer to assume that it does not and so cannot mapping the buffer on to
the CPU as a synchronisation point. The most obvious counter-example is
when we map the same buffer twice without using it in a batch.

Signed-off-by: Chris Wilson <chris@chris-wilson.co.uk>
(cherry picked from commit 637f003b047e426901cab6b1fe3a0924bcb9a38a)

Signed-off-by: Owain G. Ainsworth <oga@openbsd.org>

1 files changed, 0 insertions, 12 deletions

diff --git a/src/i830_uxa.c b/src/i830_uxa.c
index 424a0a56..2f1b48e9 100644
--- a/src/i830_uxa.c
+++ b/src/i830_uxa.c
@@ -609,18 +609,6 @@ static Bool i830_uxa_prepare_access(PixmapPtr pixmap, uxa_access_t access)
 	} else
 		i830_wait_ring_idle(scrn);
 
-	/* This acts as a synchronisation point. */
-	while (!list_is_empty(&intel->flush_pixmaps)) {
-		struct intel_pixmap *entry;
-
-		entry = list_first_entry(&intel->flush_pixmaps,
-					 struct intel_pixmap,
-					 flush);
-
-		entry->flush_read_domains = entry->flush_write_domain = 0;
-		list_del(&entry->flush);
-	}
-
 	return TRUE;
 }