逆向攻防世界CTF系列13-Reversing-x64Elf-100
无壳64位
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| __int64 __fastcall main(int a1, char **a2, char **a3) { char s[264]; unsigned __int64 v5;
v5 = __readfsqword(0x28u); printf("Enter the password: "); if ( !fgets(s, 255, stdin) ) return 0LL; if ( (unsigned int)sub_4006FD(s) ) { puts("Incorrect password!"); return 1LL; } else { puts("Nice!"); return 0LL; } }
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看看sub_4006FD
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| __int64 __fastcall sub_4006FD(__int64 a1) { int i; __int64 v3[4];
v3[0] = (__int64)"Dufhbmf"; v3[1] = (__int64)"pG`imos"; v3[2] = (__int64)"ewUglpt"; for ( i = 0; i <= 11; ++i ) { if ( *(char *)(v3[i % 3] + 2 * (i / 3)) - *(char *)(i + a1) != 1 ) return 1LL; } return 0LL; }
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看样子要求返回nice,那么sub得返回0也就是*(char *)(v3[i % 3] + 2 * (i / 3)) - *(char *)(i + a1) 都== 1
*(char *)(v3[i % 3] + 2 * (i / 3)) - 1 == *(char *)(a1+i)
结合v3[0] = (__int64)”Dufhbmf”;
可以理解为v[i%3]的第几个字符 = a1
a1+i其实就是地址,*(char *)(a1+i)就是对应地址的字符
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| v3 = ["Dufhbmf","pG`imos","ewUglpt"]
flag = ''
for i in range(0,12): flag += chr(ord(v3[i%3][2*int(i/3)]) - 1)
print(flag)
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