(v)sscanf handles %i improperly (+patch)

Vladimir Kondratiev (vladimir.kondratiev@intel.com)
Mon, 07 Jan 2002 23:27:04 +0200


Sorry for 2-nd posting, but I'm afraid 1-st one was lost. Or is the
issue itself not relevant? Anyway,

I found (v)sscanf included in last kernels handles %i format improperly.
Currently, up to 2.4.17, %i is handled identical to %d. However,
accordingly to man for sscanf,

i Matches an optionally signed integer; the next
pointer must be a pointer to int. The integer is
read in base 16 if it begins with `0x' or `0X', in
base 8 if it begins with `0', and in base 10 other­
wise. Only characters that correspond to the base
are used.

Please, when replying, CC me: mailto:vladimir.kondratiev@intel.com

Patch is quite small (against 2.4.17):

--- vsprintf.c.orig Thu Oct 11 20:17:22 2001
+++ vsprintf.c Tue Dec 25 23:29:31 2001
@@ -616,8 +616,9 @@
case 'X':
base = 16;
break;
- case 'd':
case 'i':
+ base = 0; /* autodetect */
+ case 'd':
is_sign = 1;
case 'u':
break;

-
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