memory-write

Problem

This problem comes from pwn.college: Assembly Crash Course ⤴.

We will now set some values in memory dynamically before each run. On each run, the values will change. This means you will need to perform some type of formulaic operation with registers. We will tell you which registers are set beforehand and where you should put the result. In most cases, it’s rax.

In this level, you will be working with memory. This will require you to read or write to things stored linearly in memory. If you are confused, go look at the linear addressing module in ‘ike. You may also be asked to dereference things, possibly multiple times, to things we dynamically put in memory for your use.

Recall that memory is stored linearly.

What does that mean?

Say we access the quad word at 0x1337:

[0x1337] = 0x00000000deadbeef

The real way memory is laid out is byte by byte, little endian:

[0x1337] = 0xef
[0x1337 + 1] = 0xbe
[0x1337 + 2] = 0xad
...
[0x1337 + 7] = 0x00

What does this do for us?

Well, it means that we can access things next to each other using offsets, similar to what was shown above.

Say you want the 5th byte from an address, you can access it like:

mov al, [address+4]

Remember, offsets start at 0.

Perform the following:

• Load two consecutive quad words from the address stored in rdi.
• Calculate the sum of the previous steps’ quad words.
• Store the sum at the address in rsi.

Solution

.intel_syntax noprefix
.global _start

_start:
  mov rax, [rdi]
  add rax, [rdi+8]
  mov [rsi], rax

Note: You cannot perform add [rdi], [rdi+8] as this would be a memory-to-memory operation, which is not allowed in assembly.