integer-division

Problem

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

In this level, you will be working with registers. You will be asked to modify or read from registers.

We will 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, which is usually rax.

Division in x86 is more special than in normal math. Math here is called integer math, meaning every value is a whole number.

As an example: 10 / 3 = 3 in integer math.

Why?

Because 3.33 is rounded down to an integer.

The relevant instructions for this level are:

• mov rax, reg1
• div reg2

Note: div is a special instruction that can divide a 128-bit dividend by a 64-bit divisor while storing both the quotient and the remainder, using only one register as an operand.

How does this complex div instruction work and operate on a 128-bit dividend (which is twice as large as a register)?

For the instruction div reg, the following happens:

• rax = rdx:rax / reg
• rdx = remainder

rdx:rax means that rdx will be the upper 64-bits of the 128-bit dividend and rax will be the lower 64-bits of the 128-bit dividend.

You must be careful about what is in rdx and rax before you call div.

Please compute the following:

• speed = distance / time, where:
  • distance = rdi
  • time = rsi
  • speed = rax

Note that distance will be at most a 64-bit value, so rdx should be 0 when dividing.

Solution

In assembly, the div instruction divides the rax register by either another register or a value in memory. You also cannot divide by an immediate!

The result of the division will be stored in rax.

So we must move our dividend into rax before using the div instruction with our divisor.

.intel_syntax noprefix
.global _start

_start:
  mov rax, rdi
  div rsi