wimax/i2400m: SDIO specific TX queue's minimum buffer room for new message
authorPrasanna S. Panchamukhi <prasannax.s.panchamukhi@intel.com>
Tue, 13 Apr 2010 23:36:26 +0000 (16:36 -0700)
committerInaky Perez-Gonzalez <inaky.perez-gonzalez@intel.com>
Tue, 11 May 2010 21:08:58 +0000 (14:08 -0700)
This patch specifies the TX queue's minimum buffer room required to
accommodate one smallest SDIO payload.
Please refer the documentation in the code.

Signed-off-by: Prasanna S. Panchamukhi <prasannax.s.panchamukhi@intel.com>
drivers/net/wimax/i2400m/sdio.c

index 7632f80954e3e56620b95b474ccf79d4e90037c8..9bfc26e1bc6b92facf7ab32abec51e89ce6a9980 100644 (file)
@@ -483,6 +483,13 @@ int i2400ms_probe(struct sdio_func *func,
        sdio_set_drvdata(func, i2400ms);
 
        i2400m->bus_tx_block_size = I2400MS_BLK_SIZE;
+       /*
+        * Room required in the TX queue for SDIO message to accommodate
+        * a smallest payload while allocating header space is 224 bytes,
+        * which is the smallest message size(the block size 256 bytes)
+        * minus the smallest message header size(32 bytes).
+        */
+       i2400m->bus_tx_room_min = I2400MS_BLK_SIZE - I2400M_PL_ALIGN * 2;
        i2400m->bus_pl_size_max = I2400MS_PL_SIZE_MAX;
        i2400m->bus_setup = i2400ms_bus_setup;
        i2400m->bus_dev_start = i2400ms_bus_dev_start;