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Challenge

What will asm4(“picoCTF_a3112”) return?

asm4:
    <+0>:   push   ebp
    <+1>:   mov    ebp,esp
    <+3>:   push   ebx
    <+4>:   sub    esp,0x10
    <+7>:   mov    DWORD PTR [ebp-0x10],0x246
    <+14>:  mov    DWORD PTR [ebp-0xc],0x0
    <+21>:  jmp    0x518 <asm4+27>
    <+23>:  add    DWORD PTR [ebp-0xc],0x1
    <+27>:  mov    edx,DWORD PTR [ebp-0xc]
    <+30>:  mov    eax,DWORD PTR [ebp+0x8]
    <+33>:  add    eax,edx
    <+35>:  movzx  eax,BYTE PTR [eax]
    <+38>:  test   al,al
    <+40>:  jne    0x514 <asm4+23>
    <+42>:  mov    DWORD PTR [ebp-0x8],0x1
    <+49>:  jmp    0x587 <asm4+138>
    <+51>:  mov    edx,DWORD PTR [ebp-0x8]
    <+54>:  mov    eax,DWORD PTR [ebp+0x8]
    <+57>:  add    eax,edx
    <+59>:  movzx  eax,BYTE PTR [eax]
    <+62>:  movsx  edx,al
    <+65>:  mov    eax,DWORD PTR [ebp-0x8]
    <+68>:  lea    ecx,[eax-0x1]
    <+71>:  mov    eax,DWORD PTR [ebp+0x8]
    <+74>:  add    eax,ecx
    <+76>:  movzx  eax,BYTE PTR [eax]
    <+79>:  movsx  eax,al
    <+82>:  sub    edx,eax
    <+84>:  mov    eax,edx
    <+86>:  mov    edx,eax
    <+88>:  mov    eax,DWORD PTR [ebp-0x10]
    <+91>:  lea    ebx,[edx+eax*1]
    <+94>:  mov    eax,DWORD PTR [ebp-0x8]
    <+97>:  lea    edx,[eax+0x1]
    <+100>: mov    eax,DWORD PTR [ebp+0x8]
    <+103>: add    eax,edx
    <+105>: movzx  eax,BYTE PTR [eax]
    <+108>: movsx  edx,al
    <+111>: mov    ecx,DWORD PTR [ebp-0x8]
    <+114>: mov    eax,DWORD PTR [ebp+0x8]
    <+117>: add    eax,ecx
    <+119>: movzx  eax,BYTE PTR [eax]
    <+122>: movsx  eax,al
    <+125>: sub    edx,eax
    <+127>: mov    eax,edx
    <+129>: add    eax,ebx
    <+131>: mov    DWORD PTR [ebp-0x10],eax
    <+134>: add    DWORD PTR [ebp-0x8],0x1
    <+138>: mov    eax,DWORD PTR [ebp-0xc]
    <+141>: sub    eax,0x1
    <+144>: cmp    DWORD PTR [ebp-0x8],eax
    <+147>: jl     0x530 <asm4+51>
    <+149>: mov    eax,DWORD PTR [ebp-0x10]
    <+152>: add    esp,0x10
    <+155>: pop    ebx
    <+156>: pop    ebp
    <+157>: ret    

Yes, I am not going to do this manually as I did asm3. I’m just going to convert this to nasm code and compile it. The steps I undertook were:

  1. Manually label the jumps and replace them
  2. Add the string in the data section, note this is without 0xa as many websites show. For C this is important but for assembly it isn’t. Assenbly doesn’t care about these things it just likes bits. I first included 0xa but this generates a different answer.
global _main
section .text

_main:
    push msg
    call asm4
    nop

asm4:
    push   ebp
    mov    ebp,esp
    push   ebx
    sub    esp,0x10
    mov    DWORD[ebp-0x10],0x246
    mov    DWORD[ebp-0xc],0x0
    jmp    L27
    L23:
    add    DWORD[ebp-0xc],0x1
    L27:
    mov    edx,DWORD[ebp-0xc]
    mov    eax,DWORD[ebp+0x8]
    add    eax,edx
    movzx  eax,BYTE[eax]
    test   al,al
    jne    L23
    mov    DWORD[ebp-0x8],0x1
    jmp    L138
    L51:
    mov    edx,DWORD[ebp-0x8]
    mov    eax,DWORD[ebp+0x8]
    add    eax,edx
    movzx  eax,BYTE[eax]
    movsx  edx,al
    mov    eax,DWORD[ebp-0x8]
    lea    ecx,[eax-0x1]
    mov    eax,DWORD[ebp+0x8]
    add    eax,ecx
    movzx  eax,BYTE[eax]
    movsx  eax,al
    sub    edx,eax
    mov    eax,edx
    mov    edx,eax
    mov    eax,DWORD[ebp-0x10]
    lea    ebx,[edx+eax*1]
    mov    eax,DWORD[ebp-0x8]
    lea    edx,[eax+0x1]
    mov    eax,DWORD[ebp+0x8]
    add    eax,edx
    movzx  eax,BYTE[eax]
    movsx  edx,al
    mov    ecx,DWORD[ebp-0x8]
    mov    eax,DWORD[ebp+0x8]
    add    eax,ecx
    movzx  eax,BYTE[eax]
    movsx  eax,al
    sub    edx,eax
    mov    eax,edx
    add    eax,ebx
    mov    DWORD[ebp-0x10],eax
    add    DWORD[ebp-0x8],0x1
    L138:
    mov    eax,DWORD[ebp-0xc]
    sub    eax,0x1
    cmp    DWORD[ebp-0x8],eax
    jl     L51
    mov    eax,DWORD[ebp-0x10]
    add    esp,0x10
    pop    ebx
    pop    ebp
    ret

section .data

msg db  'picoCTF_a3112'

Just like in ASM3 we just put a breakpoint at our nop instruction and read out EAX

ASM4 solution

ASM4 solution

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