13

I'm following The Security Tube’s video here.

He overviews buffer overflows, and mentions how memory is executed from highest to lowest in the stack (at least with his implementation I assume). So we pass the memory address of a function that's not called in the program, into a 3 word buffer. We overflow that buffer with a 12 character string, and then the memory address backwards. So it looks something like this:

printf "123456789abc\x32\x07\x45\xb4" | ./demo

The actual address was (b4450732).

Why is it that we display the memory address backwards, but at the end of the stack? If memory is read from high to low, shouldn't it be backwards, but at the beginning of the string we pass to the program? Obviously, this is not the case—since he showed it working. However; it seems to me that the stack would not be overflowed and the values \23x\cba987654321 would be executed.

  • 1
    "overflowed" is the word you're looking for, not "overflown". The verb form of "overflow" is a transitive verb. "To overflow" a buffer, or a bathtub. – Peter Cordes Mar 1 '15 at 18:43
  • Isn't that address b4450732? – SilverlightFox Mar 3 '15 at 11:05
  • @SilverlightFox you're right, my mistake. – theCowardlyFrench Mar 3 '15 at 13:29
16

Crash Course in Computer Architecture

In an Intel x86 and x64 architectures there is something called the stack. This is essentially where everything to determine the execution path is stored. Parameters to functions, local variables, and return addresses are all stored on the stack. CPU registers keep track of where in the stack the program is executing. You can push values and memory addresses onto the stack up to the architecture's bit-size.

What do I mean by that? If you're on a 32-bit system, then each value you push onto the stack will essentially be an unsigned int 32-bits in size (4 bytes). If you're on a 64-bit system it will be 64-bits in size (8 bytes). Below is an example[1] of what the stack looks like for a function:

uint32_t function(int a, int b, int c, int d, int e, int f, int g, int h);

enter image description here

The example above is for x64. CPU registers RDI, RSI, RDX, RCX, R8 and R9 store the first 6 parameters, and the rest are pushed on the the stack. The return address, g, and h will be 8-byte values. g might only equal 0x10, but when you push it on the stack it will look like 0x0000000000000010.

After the parameters are pushed onto the stack the call instruction is executed. This will push the return address onto the stack, and jump to the function for execution. Because the stack grows towards low address space each time you push something onto the stack it moves closer to zero. In the picture above you'll also see the local variables xx, yy, and zz. Each of these are also moving down the stack towards lower memory. Of course, realize that the program is always manipulating the top of the stack.

Stack Overflow

Lets say you create a local variable that is a buffer with a maximum of 12 bytes. Something like this, unsigned char buffer[12];. The stack makes space for 12 bytes of data. Lets say we fill this buffer with "012345678912" it's going to look like this on the stack1:

High
...
32313938
37363534
33323130
...
Low

Because the beginning of the buffer will always be towards lower memory[2]. So when you begin writing into a buffer you're always writing from Low to High. If you don't allocate enough space and you write more data than you have allocated you have a buffer overflow.

enter image description here


Endianess

Now you want to overwrite the return address on the stack. You put it at the end of your string so that as the copy writes over higher memory it overwrites your return address with the address you want there instead2. Intel throws another curve ball at you. x86 and x64 systems are Little Endian. Which means that the least significant byte is in the smallest address.

So you'll want the least significant byte of the address (0x32) written to lower memory. So you'll write the address into memory backwards because you're writing from low to high memory. When viewing the address from high to low (as the example above shows) the memory address will look correct. However, when viewing from low to high it will be backwards. The important aspect to remember is that the architecture requires the LSB to be written to lower memory.


1 - I'm using a 32-bit system here because a 4-byte width is easier to draw out.
2 - Saved EBP, I'm ignoring this because it's not important to the discussion. It's very important to remember, but just kind of ignore its existence for now.

15

There are two things going on here:

  1. On x86 and x86-64 (and most other hardware), the stack grows from the top of memory downwards. Because of this, data used by a function (eg. the buffer you're overflowing) occurs at a lower address number than data used in calling the function (eg. the address to return to after the function is done, which you're trying to overflow into).

  2. On x86 and x86-64 (and a variety of other hardware), multi-byte values such as addresses are stored in little-endian order, ie. "backwards" from the viewpoint of a person reading it.

  • 4
    I would add that \x32 is a single byte with value 32 in hexadecimal. Reading it backwards still gives \x32 and not `23x`. – user2313067 Mar 1 '15 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.