assembly_2

assembly-2

Reversing - 250 points

Challenge

What does asm2(0x7,0x28) return? Submit the flag as a hexadecimal value (starting with '0x').

Hint

assembly conditions

Solution

Assembly

asm2:
	push   	ebp
	mov    	ebp,esp
	sub    	esp,0x10
	mov    	eax,DWORD PTR [ebp+0xc]
	mov 	DWORD PTR [ebp-0x4],eax
	mov    	eax,DWORD PTR [ebp+0x8]
	mov	DWORD PTR [ebp-0x8],eax
	jmp    	part_b
part_a:	
	add    	DWORD PTR [ebp-0x4],0x1
	add	DWORD PTR [ebp+0x8],0x76
part_b:	
	cmp    	DWORD PTR [ebp+0x8],0xa1de
	jle    	part_a
	mov    	eax,DWORD PTR [ebp-0x4]
	mov	esp,ebp
	pop	ebp
	ret

Psuedocode

// => asm2(0x7,0x28)
// param2 = ebp+0xc == 0x28
// param1 = ebp+0x8 == 0x07
// return address = ebp+0x4
// base pointer = ebp

asm2:

	byte local1[16] // sub  esp,0x10 ; allocate local param		
	local1[12] = param2 // mov  DWORD PTR [ebp-0x4],eax
	local1[8] = param1 // mov	DWORD PTR [ebp-0x8],eax		
	
	while (param1 <= 0xa1de) {
		local1[12] += 1
		param1 += 0x76
	}
	return local1[12];

So the loop will run (0xa1de - 0x7) / 0x76 times or 352 times.

So local1[12] will increment 352 times to get 0x188

Flag

0x188