Use schedule_timeout_{un,}interruptible() instead of
set_current_state()/schedule_timeout() to reduce kernel size.
Signed-off-by: Nishanth Aravamudan <nacc@us.ibm.com>
Cc: Jens Axboe <axboe@suse.de>
Signed-off-by: Adrian Bunk <bunk@stusta.de>
Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>
 static void sbp_sleep(u_int time)
 {
        sti();
-       current->state = TASK_INTERRUPTIBLE;
-       schedule_timeout(time);
+       schedule_timeout_interruptible(time);
        sti();
 }
 /*==========================================================================*/
 
        /* look for the CD-ROM, follows the procedure in the DOS driver */
        inb(select_unit_reg);
        /* wait for 40 18 Hz ticks (reverse-engineered from DOS driver) */
-       set_current_state(TASK_INTERRUPTIBLE);
-       schedule_timeout((HZ+17)*40/18);
+       schedule_timeout_interruptible((HZ+17)*40/18);
        inb(result_reg);
 
        outb(0, read_status_reg);       /* does a reset? */