[Linux-parport] [patch 1/4] block/pf: replace pf_sleep() with msleep()

[domen at coderock.org]

Please consider replacing with the following patch:

Use msleep() instead of pf_sleep() 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. Remove the
definition of pf_sleep().

Signed-off-by: Nishanth Aravamudan <nacc at us.ibm.com>
Signed-off-by: Domen Puncer <domen at coderock.org>
---


 kj-domen/drivers/block/paride/pf.c |   10 ++--------
 1 files changed, 2 insertions(+), 8 deletions(-)

diff -puN drivers/block/paride/pf.c~msleep-drivers_block_paride_pf drivers/block/paride/pf.c
--- kj/drivers/block/paride/pf.c~msleep-drivers_block_paride_pf	2005-03-05 16:10:45.000000000 +0100
+++ kj-domen/drivers/block/paride/pf.c	2005-03-05 16:10:45.000000000 +0100
@@ -526,12 +526,6 @@ 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.
@@ -546,11 +540,11 @@ 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++)
_