What is a buffer overflow and how do I cause one?

问题:

I have heard about a buffer overflow and I would like to know how to cause one.

Can someone show me a small buffer overflow example? New(And what they are used for?)

回答1:

A buffer overflow is basically when a crafted section (or buffer) of memory is written outside of its intended bounds. If an attacker can manage to make this happen from outside of a program it can cause security problems as it could potentially allow them to manipulate arbitrary memory locations, although many modern operating systems protect against the worst cases of this.

While both reading and writing outside of the intended bounds are generally considered a bad idea, the term "buffer overflow" is generally reserved for writing outside the bounds, as this can cause an attacker to easily modify the way your code runs. There is a good article on Wikipedia about buffer overflows and the various ways they can be used for exploits.

In terms of how you could program one yourself, it would be a simple matter of:

char a[4];
strcpy(a,"a string longer than 4 characters"); // write past end of buffer (buffer overflow)
printf("%s\n",a[6]); // read past end of buffer (also not a good idea)

Whether that compiles and what happens when it runs would probably depend on your operating system and compiler.

回答2:

Classical example of a buffer-overflow:

// noone will ever have the time to type more than 64 characters...
char buf[64];
gets(buf); // let user put his name

The buffer overflow alone does most often not happen purposely. It happens most often because of a so-called "off-by-one" error. Meaning you have mis-calculated the array-size by one - maybe because you forgot to account for a terminating null character, or because some other stuff.

But it can also be used for some evil stuff. Indeed, the user long knew this hole, and then inserts say 70 characters, with the last ones containing some special bytes which overwrite some stack-slot - if the user is really tricky he/she will hit the return-address slot in the stack, and overwrites it so it jumps forward into that just inserted buffer: Because what the user entered was not his name, but his shell-code that he previously compiled and dumped out. That one will then just executed. There are some problems. For example, you have to arrange not to have a "\n" in that binary code (because gets would stop reading there). For other ways that mess with dangerous string functions, the binary zero is problematic because string functions stop copying there to the buffer. People have used xor with two times the same value to produce a zero too, without writing a zero byte explicitly.

That's the classic way of doing it. But there are some security blocks that can tell that such things happened and other stuff that make the stack non-executable. But i guess there are way better tricks than i just explained. Some assembler guy could probably now tell you long stories about that :)

How to avoid it

Always use functions that take a maximal-length argument too, if you are not 100% sure that a buffer is really large enough. Don't play such games as "oh, the number will not exceed 5 characters" - it will fail some day. Remember that one rocket where scientists said that the number will not exceed some magnitude, because the rocket would never be that fast. But some day, it was actually faster, and what resulted was an integer overflow and the rocket crashed (it's about a bug in Ariane 5, one of the most expensive Computer bugs in history).

For example, instead of gets use fgets. And instead of sprintf use snprintf where suitable and available (or just the C++ style things like istream and stuff)

回答3:

In the modern linux OS you can't made exploiting buffer overflow without some EXTRA experiment. why ? because you will be blocked by ASLR (Address Stack Layer Randomization) and stack protector in this modern GNU C compiler. you will not locate memory easily because memory will fall into random memory caused by ASLR. and you will blocked by stack protector if you try to overflow the program.

For begining you need to put of ASLR to be 0 default value is 2

root@bt:~# cat /proc/sys/kernel/randomize_va_space
2
root@bt:~# echo 0 > /proc/sys/kernel/randomize_va_space
root@bt:~# cat /proc/sys/kernel/randomize_va_space
0
root@bt:~#

in this is case not about OLD STYLE buffer overflow tutorial you may got from internet. or aleph one tutorial will not work anymore in your system now.

now lets make a program vulnerability to buffer overflow scenario

---------------------bof.c--------------------------
#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
        char buffer[400];
        strcpy(buffer, argv[1]);

        return 0;
}
---------------------EOF-----------------------------

looks at strcpy function is dangerous without stack protector, because function without checking how many bytes we will input. compile with extra option -fno-stack-protector dan -mpreferred-stack-boundary=2 for take off stack protector in your C program

root@bt:~# gcc -g -o bof -fno-stack-protector -mpreferred-stack-boundary=2 bof.c
root@bt:~# chown root:root bof
root@bt:~# chmod 4755 bof

buffer overflow C program with SUID root access scenatio now we have make it. now lets search how many bytes we need to put into buffer to made a program segmentation fault

root@bt:~# ./bof `perl -e 'print "A" x 400'`
root@bt:~# ./bof `perl -e 'print "A" x 403'`
root@bt:~# ./bof `perl -e 'print "A" x 404'`
Segmentation fault
root@bt:~#

you see we need 404 bytes to made program segmentation fault (crash) now how many bytes we need to overwrite EIP ? EIP is instruction will be executed after. so hacker do overwrite EIP to evil instruction what they want in the binary SUID on the program. if the program in the SUID root, the instruction will be runned in root access.

root@bt:~# gdb -q bof
(gdb) list
1       #include <stdio.h>
2       #include <string.h>
3
4       int main(int argc, char** argv)
5       {
6               char buffer[400];
7               strcpy(buffer, argv[1]);
8
9               return 0;
10      }
(gdb) run `perl -e 'print "A" x 404'`
Starting program: /root/bof `perl -e 'print "A" x 404'`

Program received signal SIGSEGV, Segmentation fault.
0xb7e86606 in __libc_start_main () from /lib/tls/i686/cmov/libc.so.6
(gdb) run `perl -e 'print "A" x 405'`
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /root/bof `perl -e 'print "A" x 405'`

Program received signal SIGSEGV, Segmentation fault.
0xb7e800a9 in ?? () from /lib/tls/i686/cmov/libc.so.6
(gdb)

program GOT segmentation fault return code. let's input more bytes and take see to EIP register.

(gdb) run `perl -e 'print "A" x 406'`
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /root/bof `perl -e 'print "A" x 406'`

Program received signal SIGSEGV, Segmentation fault.
0xb7004141 in ?? ()
(gdb)

(gdb) run `perl -e 'print "A" x 407'`
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /root/bof `perl -e 'print "A" x 407'`

Program received signal SIGSEGV, Segmentation fault.
0x00414141 in ?? ()
(gdb)

little more

(gdb) run `perl -e 'print "A" x 408'`
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /root/bof `perl -e 'print "A" x 408'`

Program received signal SIGSEGV, Segmentation fault.
0x41414141 in ?? ()
(gdb)

(gdb) i r
eax            0x0      0
ecx            0xbffff0b7       -1073745737
edx            0x199    409
ebx            0xb7fc9ff4       -1208180748
esp            0xbffff250       0xbffff250
ebp            0x41414141       0x41414141
esi            0x8048400        134513664
edi            0x8048310        134513424
eip            0x41414141       0x41414141 <-- overwriten !!
eflags         0x210246 [ PF ZF IF RF ID ]
cs             0x73     115
ss             0x7b     123
ds             0x7b     123
es             0x7b     123
fs             0x0      0
gs             0x33     51
(gdb)

now you can do your next step...

回答4:

A buffer overflow is just writing past the end of a buffer:

int main(int argc, const char* argv[])
{
    char buf[10];
    memset(buf, 0, 11);
    return 0;
}

回答5:

In addition to what has already been said, keep in mind that you'r program may or may not "crash" when a buffer overflow occurs. It should crash, and you should hope it does - but if the buffer overflow "overflows" into another address that your application has also allocated - your application may appear to operate normally for a longer period of time.

If you are using a later edition of Microsoft Visual Studio - I would suggest using the new secure counterparts in the stdlib, such as sprintf_s insted of sprintf, ect...

回答6:

This should be enought to reproduce it:

void buffer_overflow() 
{
    char * foo = "foo";
    char buffer[10];

    for(int it = 0; it < 1000; it++) {
        buffer[it] = '*';
    }

    char accessViolation = foo[0];
}

回答7:

The "classic" buffer overflow example is:

int main(int argc, char *argv[])
{
    char buffer[10];
    strcpy(buffer, argv[1]);
}

That lets you play with the buffer overflow parameters and tweak them to your hearts content. The book "Hacking - The Art of Exploitation" (Link goes to Amazon) goes into great detail about how to play around with buffer overflows (purely as an intellectual exercise obviously).

回答8:

If you want to check you program for buffer overflows, you could run it with tools like Valgrind. They will find some memory management bugs for you.

回答9:

This is a general comment about the answers you received. For example:

int main(int argc, char *argv[])
{
    char buffer[10];
    strcpy(buffer, argv[1]);
}

And:

int main(int argc, const char* argv[])
{
    char buf[10];
    memset(buf, 0, 11);
    return 0;
}

On modern Linux platforms, this may not work as expected or intended. It may not work because of the FORTIFY_SOURCE security feature.

FORTIFY_SOURCE uses "safer" variants of high risk functions like memcpy and strcpy. The compiler uses the safer variants when it can deduce the destination buffer size. If the copy would exceed the destination buffer size, then the program calls abort().

To disable FORTIFY_SOURCE for your testing, you should compile the program with -U_FORTIFY_SOURCE or -D_FORTIFY_SOURCE=0.

回答10:

In this context, a buffer is a portion of memory set aside for a particular purpose, and a buffer overflow is what happens when a write operation into the buffer keeps going past the end (writing into memory which has a different purpose). This is always a bug.

A buffer overflow attack is one which uses this bug to accomplish something that the program's author didn't intend to be possible.

回答11:

With the correct answers given: To get more into this topic, you might want to listen to the Podcast Security Now. In Episode 39 (a while back) they discussed this in depth. This is a quick way to get a deeper understanding without requiring to digest a whole book.

(At the link you'll find the archive with multiple size versions as well as a transcript, if you're rather visually oriented). Audio is not the perfect medium for this topic but Steve is working wonders to deal with this.