[PATCH] pf: replace pf_sleep() with msleep()

From: Nishanth Aravamudan <nacc@us.ibm.com>

Use msleep() instead of schedule_timeout() to guarantee the task delays
as expected. TASK_INTERRUPTIBLE is used in the original code, however
there is no check on the return values / for signals, thus I believe
TASK_UNINTERRUPTIBLE (and hence msleep()) is more appropriate.

Also, remove pf_sleep().

Signed-off-by: Nishanth Aravamudan <nacc@us.ibm.com>
Signed-off-by: Domen Puncer <domen@coderock.org>
Signed-off-by: Alexey Dobriyan <adobriyan@gmail.com>

Index: linux-kj/drivers/block/paride/pf.c
===================================================================
--- linux-kj.orig/drivers/block/paride/pf.c	2005-09-03 14:19:56.000000000 +0400
+++ linux-kj/drivers/block/paride/pf.c	2005-09-03 14:22:51.000000000 +0400
@@ -505,12 +505,6 @@ static void pf_eject(struct pf_unit *pf)
 
 #define PF_RESET_TMO   30	/* in tenths of a second */
 
-static void pf_sleep(int cs)
-{
-	current->state = TASK_INTERRUPTIBLE;
-	schedule_timeout(cs);
-}
-
 /* the ATAPI standard actually specifies the contents of all 7 registers
    after a reset, but the specification is ambiguous concerning the last
    two bytes, and different drives interpret the standard differently.
@@ -525,11 +519,11 @@ static int pf_reset(struct pf_unit *pf)
 	write_reg(pf, 6, 0xa0+0x10*pf->drive);
 	write_reg(pf, 7, 8);
 
-	pf_sleep(20 * HZ / 1000);
+	msleep(20);
 
 	k = 0;
 	while ((k++ < PF_RESET_TMO) && (status_reg(pf) & STAT_BUSY))
-		pf_sleep(HZ / 10);
+		msleep(100);
 
 	flg = 1;
 	for (i = 0; i < 5; i++)