How to look at the stack with gdb
I was chatting with someone yesterday and they mentioned that they don’t really understand exactly how the stack works or how to look at it.
So here’s a quick walkthrough of how you can use gdb to look at the stack of a C program. I think this would be similar for a Rust program, but I’m going to use C because I find it a little simpler for a toy example and also you can do Terrible Things in C more easily.
our test program#
Here’s a simple C program that declares a few variables and reads two strings from standard input. One of the strings is on the heap, and one is on the stack.
#include <stdio.h>
#include <stdlib.h>
int main() {
char stack_string[10] = "stack";
int x = 10;
char *heap_string;
heap_string = malloc(50);
printf("Enter a string for the stack: ");
gets(stack_string);
printf("Enter a string for the heap: ");
gets(heap_string);
printf("Stack string is: %s\n", stack_string);
printf("Heap string is: %s\n", heap_string);
printf("x is: %d\n", x);
}
This program uses the extremely unsafe function gets
which you should never
use, but that’s on purpose – we learn more when things go wrong.
step 0: compile the program.#
We can compile it with gcc -g -O0 test.c -o test
.
The -g
flag compiles the program with debugging symbols, which is going to
make it a lot easier to look at our variables.
O0
tells gcc to turn off optimizations which I did just to make sure ourx
variable didn’t get optimized out.
step 1: start gdb#
We can start gdb like this:
$ gdb ./test
It prints out some stuff about the GPL and then gives a prompt. Let’s create a breakpoint on the main
function.
(gdb) b main
Then we can run the program:
(gdb) b main
Starting program: /home/bork/work/homepage/test
Breakpoint 1, 0x000055555555516d in main ()
(gdb) run
Starting program: /home/bork/work/homepage/test
Breakpoint 1, main () at test.c:4
4 int main() {
Okay, great! The program is running and we can start looking at the stack
step 2: look at our variables’ addresses#
Let’s start out by learning about our variables. Each of them has an address in memory, which we can print out like this:
(gdb) p &x
$3 = (int *) 0x7fffffffe27c
(gdb) p &heap_string
$2 = (char **) 0x7fffffffe280
(gdb) p &stack_string
$4 = (char (*)[10]) 0x7fffffffe28e
So if we look at the stack at those addresses, we should be able to see all of these variables!
concept: the stack pointer#
We’re going to need to use the stack pointer so I’ll try to explain it really quickly.
There’s an x86 register called ESP called the “stack pointer”. Basically
it’s the address of the start of the stack for the current function. In gdb you can access it
with $sp
. When you call a new function or return from a function, the value
of the stack pointer changes.
step 3: look at our variables on the stack at the beginning of main
#
First, let’s look at the stack at the start of the main
function. Here’s
the value of our stack pointer right now:
(gdb) p $sp
$7 = (void *) 0x7fffffffe270
So the stack for our current function starts at 0x7fffffffe270
. Cool.
Now let’s use gdb to print out the first 40 words (aka 160 bytes) of memory after the start of the current function’s stack. It’s possible that some of this memory isn’t part of the stack because I’m not totally sure how big the stack is here. But at least the beginning of this is part of the stack.
(gdb) x/40x $sp
0x7fffffffe270: 0x00000000 0x00000000 0x555552500x00005555
0x7fffffffe280:0x00000000 0x00000000 0x55555070 0x00005555
0x7fffffffe290:0xffffe390 0x00007fff 0x00000000 0x00000000
0x7fffffffe2a0: 0x00000000 0x00000000 0xf7df4b25 0x00007fff
0x7fffffffe2b0: 0xffffe398 0x00007fff 0xf7fca000 0x00000001
0x7fffffffe2c0: 0x55555169 0x00005555 0xffffe6f9 0x00007fff
0x7fffffffe2d0: 0x55555250 0x00005555 0x3cae816d 0x8acc2837
0x7fffffffe2e0: 0x55555070 0x00005555 0x00000000 0x00000000
0x7fffffffe2f0: 0x00000000 0x00000000 0x00000000 0x00000000
0x7fffffffe300: 0xf9ce816d 0x7533d7c8 0xa91a816d 0x7533c789
I’ve bolded approximately where the stack_string
, heap_string
, and x
variables are and colour coded them:
x
is red and starts at0x7fffffffe27c
heap_string
is blue and starts at0x7fffffffe280
stack_string
is purple and starts at0x7fffffffe28e
I think I might have bolded the location of some of those variables a bit wrong here but that’s approximately where they are.
One weird thing you might notice here is that x
is the number 0x5555
, but
we set x
to 10! That because x
doesn’t actually get set until after our
main
function starts, and we’re at the very beginning of main
.
step 3: look at the stack again on line 10#
Let’s skip a few lines and wait for our variables to actually get set to the
values we initialized them to. By the time we get to line 10, x
should be set to 10.
First, we need to set another breakpoint:
(gdb) b test.c:10
Breakpoint 2 at 0x5555555551a9: file test.c, line 11.
and continue the program running:
(gdb) continue
Continuing.
Breakpoint 2, main () at test.c:11
11 printf("Enter a string for the stack: ");
Okay! Let’s look at all the same things again! gdb
is formatting the bytes in
a slightly different way here and I don’t actually know why. Here’s a reminder of where to find our variables on the stack:
x
is red and starts at0x7fffffffe27c
heap_string
is blue and starts at0x7fffffffe280
stack_string
is purple and starts at0x7fffffffe28e
(gdb) x/80x $sp
0x7fffffffe270: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0x7fffffffe278: 0x50 0x52 0x55 0x550x0a 0x00 0x00 0x00
0x7fffffffe280:0xa0 0x92 0x55 0x55 0x55 0x55 0x00 0x00
0x7fffffffe288: 0x70 0x50 0x55 0x55 0x55 0x550x73 0x74
0x7fffffffe290:0x61 0x63 0x6b 0x00 0x00 0x00 0x00 0x00
0x7fffffffe298: 0x00 0x80 0xf7 0x8a 0x8a 0xbb 0x58 0xb6
0x7fffffffe2a0: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0x7fffffffe2a8: 0x25 0x4b 0xdf 0xf7 0xff 0x7f 0x00 0x00
0x7fffffffe2b0: 0x98 0xe3 0xff 0xff 0xff 0x7f 0x00 0x00
0x7fffffffe2b8: 0x00 0xa0 0xfc 0xf7 0x01 0x00 0x00 0x00
There are a couple of interesting things to discuss here before we go further in the program.
how stack_string
is represented in memory#
Right now (on line 10) stack_string
is set to “stack”. Let’s take a look at
how that’s represented in memory.
We can print out the bytes in the string like this:
(gdb) x/10x stack_string
0x7fffffffe28e: 0x73 0x74 0x61 0x63 0x6b 0x00 0x00 0x00
0x7fffffffe296: 0x00 0x00
The string “stack” is 5 characters which corresponds to 5 ASCII bytes –
0x73
, 0x74
, 0x61
, 0x63
, and 0x6b
. 0x73
is s
in ASCII, 0x74
is
t
, etc.
We can also get gdb to show us the string with x/1s
:
(gdb) x/1s stack_string
0x7fffffffe28e: "stack"
how heap_string
and stack_string
are different#
You’ll notice that stack_string
and heap_string
are represented in very
different ways on the stack:
stack_string
has the contents of the string (“stack”)heap_string
is a pointer to an address somewhere else in memory
Here are the bytes on the stack for the heap_string
variable:
0xa0 0x92 0x55 0x55 0x55 0x55 0x00 0x00
These bytes actually get read backwards because x86 is little-endian, so the
memory address of heap_string
is 0x5555555592a0
Another way to see the address of heap_string
in gdb is just to print it out
with p
:
(gdb) p heap_string
$6 = 0x5555555592a0 ""
the bytes that represent the integer x
#
x
is a 32-bit integer, and the bytes that represent it are 0x0a 0x00 0x00 0x00
.
We need to read these bytes backwards again (the same way reason we read the
bytes for heap_string
address backwards), so this corresponds to the number
0x000000000a
, or 0xa
, which is 10.
That makes sense! We set int x = 10;
!
step 4: read input from standard input#
Okay, we’ve initialized the variables, now let’s see how the stack changes when this part of the C program runs:
printf("Enter a string for the stack: ");
gets(stack_string);
printf("Enter a string for the heap: ");
gets(heap_string);
We need to set another breakpoint:
(gdb) b test.c:16
Breakpoint 3 at 0x555555555205: file test.c, line 16.
and continue running the program
(gdb) continue
Continuing.
We’re prompted for 2 strings, and I entered 123456789012
for the stack string
and bananas
for the heap.
let’s look at stack_string
first (there’s a buffer overflow!)#
(gdb) x/1s stack_string
0x7fffffffe28e: "123456789012"
That seems pretty normal, right? We entered 123456789012
and now it’s set to 123456789012
.
But there’s something weird about this. Here’s what those bytes look like on the stack. They’re highlighted in purple again.
0x7fffffffe270: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0x7fffffffe278: 0x50 0x52 0x55 0x55 0x0a 0x00 0x00 0x00
0x7fffffffe280: 0xa0 0x92 0x55 0x55 0x55 0x55 0x00 0x00
0x7fffffffe288: 0x70 0x50 0x55 0x55 0x55 0x550x31 0x32
0x7fffffffe290:0x33 0x34 0x35 0x36 0x37 0x38 0x39 0x30
0x7fffffffe298:0x31 0x32 0x00 0x8a 0x8a 0xbb 0x58 0xb6
0x7fffffffe2a0: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0x7fffffffe2a8: 0x25 0x4b 0xdf 0xf7 0xff 0x7f 0x00 0x00
0x7fffffffe2b0: 0x98 0xe3 0xff 0xff 0xff 0x7f 0x00 0x00
0x7fffffffe2b8: 0x00 0xa0 0xfc 0xf7 0x01 0x00 0x00 0x00
The weird thing about this is that stack_string was only supposed to be 10 bytes. But now suddenly we’ve put 13 bytes in it? What’s happening?
This is a classic buffer overflow, and what’s happening is that stack_string
wrote over other data from the program. This hasn’t caused a problem yet in our
case, but it can crash your program or, worse, open you up to Very Bad Security
Problems.
For example, if stack_string
were before heap_string
in memory, then we
could overwrite the address that heap_string
points to. I’m not sure exactly
what’s in memory after stack_string
here but we could probably use this to do
some kind of shenanigans.
something actually detects the buffer overflow#
When I cause this buffer overflow problem, here’s
./test
Enter a string for the stack: 01234567891324143
Enter a string for the heap: adsf
Stack string is: 01234567891324143
Heap string is: adsf
x is: 10
*** stack smashing detected ***: terminated
fish: Job 1, './test' terminated by signal SIGABRT (Abort)
My guess about what’s happening here is that the stack_string
variable is
actually at the end of this function’s stack, and so the extra bytes are going into a
different region of memory.
When you do this intentionally as a security exploit it’s called “stack smashing”, and somehow something is detecting that this is happening. I’m not totally sure how this is being detected.
I also thing this is interesting because the program gets killed, but it doesn’t get killed immediately when the buffer overflow happens – a few more lines of code run after the buffer overflow and the program gets killed later. Weird!
That’s all I have to say about buffer overflows.
now let’s look at heap_string
#
We also read a value (bananas
) into the heap_string
variable. Let’s see what that
looks like in memory.
Here’s what heap_string
looks on the stack after we read the variable in.
(gdb) x/40x $sp
0x7fffffffe270: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0x7fffffffe278: 0x50 0x52 0x55 0x55 0x0a 0x00 0x00 0x00
0x7fffffffe280:0xa0 0x92 0x55 0x55 0x55 0x55 0x00 0x00
0x7fffffffe288: 0x70 0x50 0x55 0x55 0x55 0x55 0x31 0x32
0x7fffffffe290: 0x33 0x34 0x35 0x36 0x37 0x38 0x39 0x30
The thing to notice here is that it looks exactly the same! It’s an address, and the address hasn’t changed. But let’s look at what’s at that address.
(gdb) x/10x 0x5555555592a0
0x5555555592a0: 0x62 0x61 0x6e 0x61 0x6e 0x61 0x73 0x00
0x5555555592a8: 0x00 0x00
Those are the bytes for bananas
! Those bytes aren’t in the stack at all,
they’re somewhere else in memory (on the heap)