snprintf((char *) ?, 0, ...); always returns Zero and doesn't change the data.
Thus the execution of
snprintf(NULL, 0, "[%5lu.000000] ", (unsigned long)ts);
has no effect on program.
The substitution with 0 increases the readability of the code.
Signed-off-by: Sebastian Duda <[email protected]>
Signed-off-by: Tobias Baumeister <[email protected]>
---
kernel/printk/printk.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/kernel/printk/printk.c b/kernel/printk/printk.c
index 5028f4f..fe3fec1 100644
--- a/kernel/printk/printk.c
+++ b/kernel/printk/printk.c
@@ -1186,7 +1186,7 @@ static size_t print_time(u64 ts, char *buf)
rem_nsec = do_div(ts, 1000000000);
if (!buf)
- return snprintf(NULL, 0, "[%5lu.000000] ", (unsigned long)ts);
+ return 0;
return sprintf(buf, "[%5lu.%06lu] ",
(unsigned long)ts, rem_nsec / 1000);
--
2.7.4
