PicoCTF - asm4

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PicoCTF - asm4

Description

What will asm4(“picoCTF_f97bb”) 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.

Source code

:::spoiler Source Code

asm4:
	<+0>:	push   ebp
	<+1>:	mov    ebp,esp
	<+3>:	push   ebx
	<+4>:	sub    esp,0x10
	<+7>:	mov    DWORD PTR [ebp-0x10],0x27a
	<+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

:::

Recon

這一題和前幾個系列的題目差不多，只是變得更臭更長，一直在找能夠動態執行的工具或腳本，不過忍不住看了1的script後，才發現自己幹一個比較快

Exploit

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#include <stdio.h>
#include <stdlib.h>

int asm4(char* in)
{
    int val;

    asm (
        "nop;"
        "nop;"
        "nop;"
        //"push   ebp;"
        //"mov    ebp,esp;"
        "push   ebx;"
        "sub    esp,0x10;"
        "mov    DWORD PTR [ebp-0x10],0x27a;"
        "mov    DWORD PTR [ebp-0xc],0x0;"
        "jmp    asm4_27;"
    "asm4_23:"
        "add    DWORD PTR [ebp-0xc],0x1;"
    "asm4_27:"
        "mov    edx,DWORD PTR [ebp-0xc];"
        "mov    eax,DWORD PTR [%[pInput]];"
        "add    eax,edx;"
        "movzx  eax,BYTE PTR [eax];"
        "test   al,al;"
        "jne    asm4_23;"
        "mov    DWORD PTR [ebp-0x8],0x1;"
        "jmp    asm4_138;"
    "asm4_51:"
        "mov    edx,DWORD PTR [ebp-0x8];"
        "mov    eax,DWORD PTR [%[pInput]];"
        "add    eax,edx;"
        "movzx  eax,BYTE PTR [eax];"
        "movsx  edx,al;"
        "mov    eax,DWORD PTR [ebp-0x8];"
        "lea    ecx,[eax-0x1];"
        "mov    eax,DWORD PTR [%[pInput]];"
        "add    eax,ecx;"
        "movzx  eax,BYTE PTR [eax];"
        "movsx  eax,al;"
        "sub    edx,eax;"
        "mov    eax,edx;"
        "mov    edx,eax;"
        "mov    eax,DWORD PTR [ebp-0x10];"
        "lea    ebx,[edx+eax*1];"
        "mov    eax,DWORD PTR [ebp-0x8];"
        "lea    edx,[eax+0x1];"
        "mov    eax,DWORD PTR [%[pInput]];"
        "add    eax,edx;"
        "movzx  eax,BYTE PTR [eax];"
        "movsx  edx,al;"
        "mov    ecx,DWORD PTR [ebp-0x8];"
        "mov    eax,DWORD PTR [%[pInput]];"
        "add    eax,ecx;"
        "movzx  eax,BYTE PTR [eax];"
        "movsx  eax,al;"
        "sub    edx,eax;"
        "mov    eax,edx;"
        "add    eax,ebx;"
        "mov    DWORD PTR [ebp-0x10],eax;"
        "add    DWORD PTR [ebp-0x8],0x1;"
    "asm4_138:"
        "mov    eax,DWORD PTR [ebp-0xc];"
        "sub    eax,0x1;"
        "cmp    DWORD PTR [ebp-0x8],eax;"
        "jl     asm4_51;"
        "mov    eax,DWORD PTR [ebp-0x10];"
        "add    esp,0x10;"
        "pop    ebx;"
        //"pop    ebp;"
        //"ret    ;"
        "nop;"
        "nop;"
        "nop;"
            :"=r"(val)
            : [pInput] "m"(in)
    );
    
    return val;
}

int main(int argc, char** argv)
{
    printf("0x%x\n", asm4("picoCTF_f97bb"));
    
    return 0;
}

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$ gcc -masm=intel -m32 exp.c -o exp
$ ./exp
0x265

Reference

  1. [asm4](https://github.com/Dvd848/CTFs/blob/master/2019_picoCTF/asm4.md ↩