Vulnerability Development mailing list archives

partial analysis of vulndev-1.c


From: "David R. Piegdon" <fleshyCPU () gmx net>
Date: Tue, 13 May 2003 20:35:06 +0200

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hope its ok that i answer to the list :)


first thing :)
as far as i know, in PLAIN C a function call is not allowed
during the definition of a variable
{ char    *buf1 = malloc(SIZE); }
but in C++, it is allowed.
but actually i am not that sure, because gcc simply compiles it
without a warning. someone knows? i've lent my favorite C book
to a friend.

second thing:
two ways are used to copy a string into a buffer:
[1]  {  strncpy(buf2, p2, SIZE);  }
and
[2]  {  for (i = 0; i <= SIZE && p1[i] != '\0'; i++) buf1[i] = p1[i];   }

it is obvious that [1] will always copy SIZE chars.
thats bad because in this case, if strlen(p2) >= SIZE, the final string
in buf1 will NOT be terminated with a NULL.
this could be used later on for something bad. but not in this context.

in opposite, [2] will copy MIN( 0..SIZE , STRLEN(p1)+1 ),
that is: MIN( SIZE+1 , STRLEN(p1)+1 ), which will write one char beyond
the end of the buffer, if the string p1 is longer or equal SIZE.
actually, this string will NEVER be null-terminated
(just look, when the for-loop is terminated: if it finds a \000 char)


so here we have two bad things:
1. strings that are not null-terminated (may not be too bad, if handled
properly later on when playing with the strings)
2. a buffer overflow, if strlen(p1) >= SIZE.
actually my gcc-compiled version does not catch this one with a SIGSEGV,
but i don't know why. i've checked it with ddd, it really overwrites one
char behind the end of the buffer.

now the question: can we use this buffer overflow?
actually in this case not, because the allocation of the buffer is done
with malloc. on linux at least :) malloc does not use the stack but it
uses the HEAP.
- ------------------------------------------------------------------------ >8
 // vulndev-1.c
 // vuln-dev mailing list security challenge #1
 // by Aaron Adams <aadams () securityfocus com>
 // Spot the error in this program.

 #include <stdio.h>
 #include <stdlib.h>

 #define SIZE    252

 int
 main(int argc, char *argv[])
 {
         int     i;
         char    *p1, *p2;
         char    *buf1 = malloc(SIZE);
         char    *buf2 = malloc(SIZE);

         if (argc != 3)
                 exit(1);

         p1 = argv[1], p2 = argv[2];
         strncpy(buf2, p2, SIZE);
         for (i = 0; i <= SIZE && p1[i] != '\0'; i++)
                 buf1[i] = p1[i];

         free(buf1);
         free(buf2);

         return 0;
 }
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