Commit 83ff56f4 authored Mar 09, 2010 by Masami Hiramatsu Committed by Ingo Molnar Mar 11, 2010

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>

parent a12b51c4

kernel/kprobes.c

+2 −1

Original line number	Diff line number	Diff line
		@@ -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) {