From: Borislav Petkov <[email protected]> If a call to panic() terminates the string with a \n, the result puts the closing brace ']---' on a newline because panic() itself adds \n too.
Now, if one goes and removes the newline chars from all panic() invocations - and the stats right now look like this: ~300 calls with an \n ~500 calls without a \n one is destined to a neverending game of whack-a-mole because the usual thing to do is add a newline at the end of a string a function is supposed to print. Therefore, simply zap any \n at the end of the panic string to avoid touching so many places in the kernel. Signed-off-by: Borislav Petkov <[email protected]> Cc: Andrew Morton <[email protected]> Cc: Kees Cook <[email protected]> Cc: "Steven Rostedt (VMware)" <[email protected]> Cc: [email protected] --- kernel/panic.c | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/kernel/panic.c b/kernel/panic.c index 8b2e002d52eb..5776d2879650 100644 --- a/kernel/panic.c +++ b/kernel/panic.c @@ -136,7 +136,7 @@ void panic(const char *fmt, ...) { static char buf[1024]; va_list args; - long i, i_next = 0; + long i, i_next = 0, len; int state = 0; int old_cpu, this_cpu; bool _crash_kexec_post_notifiers = crash_kexec_post_notifiers; @@ -173,8 +173,12 @@ void panic(const char *fmt, ...) console_verbose(); bust_spinlocks(1); va_start(args, fmt); - vsnprintf(buf, sizeof(buf), fmt, args); + len = vscnprintf(buf, sizeof(buf), fmt, args); va_end(args); + + if (len && buf[len - 1] == '\n') + buf[len - 1] = '\0'; + pr_emerg("Kernel panic - not syncing: %s\n", buf); #ifdef CONFIG_DEBUG_BUGVERBOSE /* -- 2.19.0.271.gfe8321ec057f
