picoCTF - asm2 writeup

30 Nov, 2020

Description

What does asm2(0x4,0x21) return? Submit the flag as a hexadecimal value (starting with ‘0x’). NOTE: Your submission for this question will NOT be in the normal flag format.

Category: Reverse Engineering

Points: 250

Source Code

I again added my own comments:

# asm2(0x4, 0x21) --> asm2(4, 33)

asm2:
	<+0>:	push   ebp				; set up stack frame
	<+1>:	mov    ebp,esp
	<+3>:	sub    esp,0x10				; alloc local variables
	var1 and var2
	<+6>:	mov    eax,DWORD PTR [ebp+0xc]		; eax  = 33
	<+9>:	mov    DWORD PTR [ebp-0x4],eax		; var1 = 33
	<+12>:	mov    eax,DWORD PTR [ebp+0x8]		; eax  = 4
	<+15>:	mov    DWORD PTR [ebp-0x8],eax		; var2 = 4
	<+18>:	jmp    0x509 <asm2+28>			; unconditional jump
	<+20>:	add    DWORD PTR [ebp-0x4],0x1		; var1 += 1
	<+24>:	add    DWORD PTR [ebp-0x8],0x74		; var2 += 116
	<+28>:	cmp    DWORD PTR [ebp-0x8],0xfb46	; cmp 4 with 64326
	<+35>:	jle    0x501 <asm2+20>			; loop condition
	<+37>:	mov    eax,DWORD PTR [ebp-0x4]		; eax = var1
	<+40>:	leave					; clean up stack frame
	<+41>:	ret

Solution

Option 1 - Quick Math

Two local variables are created and initialized to the function parameters.

The two variables are incremented in a loop until var2 is greater than 0xfb46 (64326). var1 is incremented by 1 and var2 is incremented by 116.

Some quick math shows us that 64326 / 116 = 554. Add 1 to this for the initial loop check to get a total of 555 loop iterations.

So var1 (which was initially 33) is now 33 + (1 * 555) = 588 –> 0x24c

Option 2 - Python

I also created a Python script to fact check my math:

# flag.py
var1  = 33
var2  = 4
num   = 64326
inc   = 116
iters = 0

# count total number of loop iterations
while var2 < num:
    iters += 1
    var2  += inc


flag = var1 + iters
flag = hex(flag)

print("The flag is {}".format(flag))

Run it:

jason@lapras:~/ctf-solutions/picoctf/reverse-engineering/asm2$ python flag.py
The flag is 0x24c