On Wed, 2012-06-06 at 02:28 +0200, Kay Sievers wrote:
On Wed, Jun 6, 2012 at 2:19 AM, Joe Perches joe@perches.com wrote:
On Wed, 2012-06-06 at 02:13 +0200, Kay Sievers wrote:
The question is what happens if you inject your new binary two-byte prefix, like: echo -e "\x01\x02Hello" > /dev/kmsg
It's not a 2 byte binary. It's a leading ascii SOH and a standard ascii char '0' ... '7' or 'd'.
#define KERN_EMERG KERN_SOH "0" /* system is unusable */ #define KERN_ALERT KERN_SOH "1" /* action must be taken immediately */ etc...
Ok.
And if that changes the log-level to "2" instead of the default "4"?
No it doesn't.
So: echo -e "\x012Hello" > /dev/kmsg is still level 4? Sounds all fine then.
Yes. # echo -e "\x012Hello again Kay" > /dev/kmsg gives: 12,780,6031964979;Hello again Kay
It's not triggering that because devkmsg_writev does prefix parsing only on the old "<n>" form.
Yeah, but printk_emit() will not try to parse it? I did not check, but with your change, the prefix parsing in printk_emit() is still skipped if a level is given as a parameter to printk_emit(), right?
If level is not -1, then whatever prefix level the string has is ignored by vprintk_emit.
from vprintk_emit:
/* strip syslog prefix and extract log level or control flags */ kern_level = printk_get_level(text); if (kern_level) { const char *end_of_header = printk_skip_level(text); switch (kern_level) { case '0' ... '7': if (level == -1) level = kern_level - '0'; case 'd': /* Strip d KERN_DEFAULT, start new line */ plen = 0; default: if (!new_text_line) { log_buf_emit_char('\n'); new_text_line = 1; } } text_len -= end_of_header - text; text = (char *)end_of_header; }
Only level == -1 will use the prefix level.
devkmsg_writev always passes a non -1 level.
cheers, Joe