1 files changed, 14 insertions, 7 deletions

diff --git a/drivers/gpu/nvgpu/common/fifo/channel.c b/drivers/gpu/nvgpu/common/fifo/channel.c
index 8af5f5e9..54fa4747 100644
--- a/drivers/gpu/nvgpu/common/fifo/channel.c
+++ b/drivers/gpu/nvgpu/common/fifo/channel.c
@@ -753,16 +753,23 @@ static int channel_gk20a_alloc_priv_cmdbuf(struct channel_gk20a *c,
          * Any sync-pt fences will take less memory so we can ignore them for
          * now.
          *
-         * A semaphore ACQ (fence-wait) is 8 dwords: semaphore_a, semaphore_b,
+         * A semaphore ACQ (fence-wait) is 8 words: semaphore_a, semaphore_b,
          * semaphore_c, and semaphore_d. A semaphore INCR (fence-get) will be 10
-         * dwords: all the same as an ACQ plus a non-stalling intr which is
-         * another 2 dwords.
+         * words: all the same as an ACQ plus a non-stalling intr which is
+         * another 2 words.
          *
-         * Lastly the number of gpfifo entries per channel is fixed so at most
-         * we can use 2/3rds of the gpfifo entries (1 pre-fence entry, one
-         * userspace entry, and one post-fence entry). Thus the computation is:
+         * We have two cases to consider: the first is we base the size of the
+         * priv_cmd_buf on the gpfifo count. Here we multiply by a factor of
+         * 2/3rds because only at most 2/3rds of the GPFIFO can be used for
+         * sync commands:
          *
-         *   (gpfifo entry number * (2 / 3) * (8 + 10) * 4 bytes.
+         *   nr_gpfifos * (2 / 3) * (8 + 10) * 4 bytes
+         *
+         * If instead num_in_flight is specified then we will use that to size
+         * the priv_cmd_buf. The worst case is two sync commands (one ACQ and
+         * one INCR) per submit so we have a priv_cmd_buf size of:
+         *
+         *   num_in_flight * (8 + 10) * 4 bytes
          */
         size = num_in_flight * 18U * (u32)sizeof(u32);
         if (gpfifo_based) {