On Wed, 2012-06-06 at 02:28 +0200, Kay Sievers wrote:
> On Wed, Jun 6, 2012 at 2:19 AM, Joe Perches <j...@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
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