On 4/23/2018 8:58 PM, Rohit Zambre wrote:
The calculation of need_uuar_lock rightly accounts for at least 1 medium
bfreg.
The code should support even 0 medium bfregs comparing the suggested
patch below.
However, the user can wrongly request for all but one uuars to be
low latency uuars.
For example, setting MLX5_TOTAL_UUARS=16 and MLX5_NUM_LOW_LAT_UUARS=15
doesn't throw any error but the user gets only 14 low latency uuars;
need_lock for bf_1 evaluates to 1. In this case, the first QP of the
context is mapped to bf_1 which is not a low latency uuar.
Signed-off-by: Rohit Zambre <rzambre@xxxxxxx>
---
providers/mlx5/mlx5.c | 3 ++-
1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/providers/mlx5/mlx5.c b/providers/mlx5/mlx5.c
index 5590241..2a09a95 100644
--- a/providers/mlx5/mlx5.c
+++ b/providers/mlx5/mlx5.c
@@ -1026,7 +1026,8 @@ static struct verbs_context *mlx5_alloc_context(struct ibv_device *ibdev,
goto err_free;
}
- if (low_lat_uuars > tot_uuars - 1) {
+ /* account for at least 1 med bfreg and the 0th uuar */
+ if (low_lat_uuars > tot_uuars - 2) {
The expected fix should not be here but in need_uuar_lock. Please see
below a candidate suggestion from Eli (cc).
errno = ENOMEM;
goto err_free;
}
static int need_uuar_lock(struct mlx5_context *ctx, int uuarn)
{
+ int i;
+
if (uuarn == 0 || mlx5_single_threaded)
return 0;
- if (uuarn >= (ctx->tot_uuars - ctx->low_lat_uuars) * 2)
+ i = (uuarn / 2) + (uuarn % 2);
+ if (i >= ctx->tot_uuars - ctx->low_lat_uuars)
return 0;
Can you please test and confirm whether it works for you ?
Yishai
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