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:

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On Wed, 2012-06-06 at 02:13 +0200, Kay Sievers wrote:

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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.

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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

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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