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sysrq: Use panic() to force a crash 

sysrq_handle_crash() currently forces a crash by dereferencing a
NULL pointer, which is undefined behavior in C. Just call panic()
instead, which is simpler and doesn't depend on compiler specific
handling of the undefined behavior.

Remove the comment on why the RCU lock needs to be released, it isn't
accurate anymore since the crash now isn't handled by the page fault
handler (for reference: the comment was added by commit 984cf355ae
("sysrq: Fix warning in sysrq generated crash.")). Releasing the lock
is still good practice though.

Suggested-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>
Signed-off-by: Matthias Kaehlcke <mka@chromium.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>
hifive-unleashed-5.1
Matthias Kaehlcke 2018-09-20 10:12:53 -07:00 committed by Greg Kroah-Hartman
parent 279070b96a
commit 8341f2f222
1 changed files with 3 additions and 10 deletions
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13
drivers/tty/sysrq.c
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@ -134,17 +134,10 @@ static struct sysrq_key_op sysrq_unraw_op = {
static void sysrq_handle_crash(int key)
{
char *killer = NULL;
/* we need to release the RCU read lock here,
* otherwise we get an annoying
* 'BUG: sleeping function called from invalid context'
* complaint from the kernel before the panic.
*/
/* release the RCU read lock before crashing */
rcu_read_unlock();
panic_on_oops = 1; /* force panic */
wmb();
*killer = 1;
panic("sysrq triggered crash\n");
}
static struct sysrq_key_op sysrq_crash_op = {
.handler = sysrq_handle_crash,