x86: Don't check for BIOS corruption in first 64K when there's no need to

Due to commit 781c5a67f1 it is
likely that the number of areas to scan for BIOS corruption is 0
 -- especially when the first 64K is already reserved
(X86_RESERVE_LOW is 64K by default).

If that's the case then don't set up the scan.

Signed-off-by: Naga Chumbalkar <nagananda.chumbalkar@hp.com>
Cc: <stable@kernel.org>
LKML-Reference: <20110225202838.2229.71011.sendpatchset@nchumbalkar.americas.hpqcorp.net>
Signed-off-by: Ingo Molnar <mingo@elte.hu>
This commit is contained in:
Naga Chumbalkar 2011-02-25 20:31:55 +00:00 committed by Ingo Molnar
parent 5471262290
commit a7bd1dafdc
1 changed files with 4 additions and 4 deletions

View File

@ -106,8 +106,8 @@ void __init setup_bios_corruption_check(void)
addr += size;
}
printk(KERN_INFO "Scanning %d areas for low memory corruption\n",
num_scan_areas);
if (num_scan_areas)
printk(KERN_INFO "Scanning %d areas for low memory corruption\n", num_scan_areas);
}
@ -143,12 +143,12 @@ static void check_corruption(struct work_struct *dummy)
{
check_for_bios_corruption();
schedule_delayed_work(&bios_check_work,
round_jiffies_relative(corruption_check_period*HZ));
round_jiffies_relative(corruption_check_period*HZ));
}
static int start_periodic_check_for_corruption(void)
{
if (!memory_corruption_check || corruption_check_period == 0)
if (!num_scan_areas || !memory_corruption_check || corruption_check_period == 0)
return 0;
printk(KERN_INFO "Scanning for low memory corruption every %d seconds\n",