Win95_OEMKey_Check.c

/*
	原逆向作者/Original Reverser：nezza@github
	来自于/From：https://gist.github.com/nezza/a25bee13f25a1733a4c7a1d3d1cf5882
	
	本人将所有的十六进制转换成十进制，并重写了部分代码，加上了详细的中英注释
	I converted all Hex into Dec, rewrote some function, and wrote detailed comments in Chinese and English.
	
	本代码以The Unlicense协议发布，详细文本可在 http://unlicense.org/ 中查看
	These code are redistributed under The Unlicense, which you could check them at http://unlicense.org/.
	
	数组计数从0开始，请务必注意
	Note that array in C counts from 0.
	
	如果你能告诉我英语部分的错误，我将会十分高兴的
	If you could tell something about my error in my English part, I would be glad.
	
	2021 SuperBart & BenderBolg
*/

undefined2 __stdcall16far check_oem_key(char *key,int len)
{
	int first_3_digits_int;		 //整型变量：前三位数字	You must know
	int second_2_digits_int;	 //整型变量：之后两位数字                what they can do
	char first_three_digits[4];  	 //字符串：前三位数字                                    from the code ;-)
	char second_two_digits[3];	 //字符串：之后两位数字                                                     Right?
	undefined2 check_result;	 //应为原程序中的结构体：是否有效  It should be a struct in the origianl code, which marks the keys is vaild or not.

	/* 密钥长度为23 第6,10&18位为'-'字符   The length of the key is 23. The 6th, 10th & 18th charactor must be charactor '-'. */
	if ((strlen(key) == 23) && (key[5] == '-') && (key[9] == '-') && (key[17] == '-'))
	{
		/* 并且是OEM密钥格式    And shall be same of the OEM key structer */
		if ( check_oem_str(3,0x11704f9d,key + 6,len) == 0 )
		{
			/* 原逆向者注释：检查前五个字符，必须是数字  From the original reverser: Check 5 charactors at the beginning, make sure they are all 'nunbers'. */
 			int i;		//整型变量：循环使用的数字  A integer variable used in loop.
			for (i = 0; i < 5; ++i)
				if (key[i] < '0' && key[i] > '9')
					return 0;
			
			/* 以下代码是将密钥前三个字符提取出来，转换为数字 Take first 3 charactors out, and translated them into integer. */
			copy_partial(3,key,len,first_three_digits);
			first_3_digits_int = atoi(first_three_digits);
			
			/* 判断前三个数字是否在0~366的闭区间范围中，若是，对后两位代码进行同样处理 */
			/* Check the first 3 charactors is between 0 and 366. If it is, do the same thing at the later charactor. */
			if ((first_3_digits_int != 0) && (first_3_digits_int < 367))
			{
				copy_partial(2,key + 3,len,second_two_digits);
				second_2_digits_int = atoi(second_two_digits);
				
				/* 判断后两个数字不在03~94的闭区间范围内 */
				/* Check the later 2 charactors, it should not between 03 and 94.  */
				/* 判断第11位是0，第17位是在0~8的开区间内 */
				/* The 11th charactor is '0', and the 17th charactor should between '0' and '8'. */
				if ((second_2_digits_int < 3) || (94 < second_2_digits_int))
				{
					if ((key[10] == '0') && ('0' < key[16]) && (key[16] < '8'))
					{
						/* 保证后面五位全都是数字字符 */
						/* Make sure the last 5 charactors are 'numbers' */
						for (i = 18; i < 23; ++i)
							if ((key[i] < '0') && (key[i] > '9'))
								return 0;
						
						/* 现在只要对中间的七位进行相加，并除以7取余。详情请看视频 bilibili.com/video/BV1Kp4y1p7LF，空降 05:08 */
						/* Add the middle 7 charactors, then mod by 7. For more details, see 05:08 at youtube.com/watch?v=cwyH59nACzQ */
						if (i == 22)
						{
							check_result = mod7_check(key + 10,len);
							return check_result;
						}
					}
				}
			}
		}
	}
}