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kprobes: Calculate the index correctly when freeing the out-of-line execution slot 

From : Ananth N Mavinakayanahalli <ananth@in.ibm.com>

When freeing the instruction slot, the arithmetic to calculate
the index of the slot in the page needs to account for the total
size of the instruction on the various architectures.

Calculate the index correctly when freeing the out-of-line
execution slot.

Reported-by: Sachin Sant <sachinp@in.ibm.com>
Reported-by: Heiko Carstens <heiko.carstens@de.ibm.com>
Signed-off-by: Ananth N Mavinakayanahalli <ananth@in.ibm.com>
Signed-off-by: Masami Hiramatsu <mhiramat@redhat.com>
LKML-Reference: <4B9667AB.9050507@redhat.com>
Signed-off-by: Ingo Molnar <mingo@elte.hu>
This commit is contained in:
Masami Hiramatsu 2010-03-09 10:22:19 -05:00 committed by Ingo Molnar
parent a12b51c478
commit 83ff56f46a
1 changed files with 2 additions and 1 deletions
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3
kernel/kprobes.c
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@ -259,7 +259,8 @@ static void __kprobes __free_insn_slot(struct kprobe_insn_cache *c,
struct kprobe_insn_page *kip;
list_for_each_entry(kip, &c->pages, list) {
long idx = ((long)slot - (long)kip->insns) / c->insn_size;
long idx = ((long)slot - (long)kip->insns) /
(c->insn_size * sizeof(kprobe_opcode_t));
if (idx >= 0 && idx < slots_per_page(c)) {
WARN_ON(kip->slot_used[idx] != SLOT_USED);
if (dirty) {