[原创]BUUCTF逆向题：[UTCTF2020]basic-re-CTF对抗-看雪-安全社区|安全招聘|kanxue.com

[原创]BUUCTF逆向题：[UTCTF2020]basic-re

发表于: 2022-3-27 17:36 4889

[原创]BUUCTF逆向题：[UTCTF2020]basic-re

宇宙大魔王

2022-3-27 17:36

4889

1.基本信息探查：

1.EXEinfo：

2.运行一下：

2.IDA分析：

1.主函数分析：

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char v4; // [rsp+Fh] [rbp-21h] BYREF
  double v5; // [rsp+10h] [rbp-20h] BYREF
  double v6; // [rsp+18h] [rbp-18h] BYREF
  unsigned __int64 v7; // [rsp+28h] [rbp-8h]
 
  v7 = __readfsqword(0x28u);
  printf("Enter an operator (+, -, *,): ");
  __isoc99_scanf("%c", &v4);
  printf("Enter two operands: ");
  __isoc99_scanf("%lf %lf", &v5, &v6);
  if ( v4 == '/' )
  {
    printf("%.1lf / %.1lf = %.1lf", v5, v6, v5 / v6);
    return 0;
  }
  if ( v4 > '/' )
    goto LABEL_11;
  if ( v4 == '-' )
  {
    printf("%.1lf - %.1lf = %.1lf", v5, v6, v5 - v6);
    return 0;
  }
  if ( v4 > '-' )
    goto LABEL_11;
  if ( v4 == '*' )
  {
    printf("%.1lf * %.1lf = %.1lf", v5, v6, v5 * v6);
    return 0;
  }
  if ( v4 != '+' )
  {
LABEL_11:
    printf("Error! operator is not correct");
    return 0;
  }
  printf("%.1lf + %.1lf = %.1lf", v5, v6, v5 + v6);
  return 0;
}