Question

我正在做一个作业，我必须通过在要编码/解码的字符串上添加/减去“密码”（另一个字符串）来“编码”和“解码”字符串。但是，当我运行它时，我得到的结果与正确的输出类似，但仍然不正确。我认为这可能与我添加/减去两个字符串的方式有关。抱歉，这很琐碎，但我是初学者。我将不胜感激！

这是用于编码和解码的代码（注意：我还必须编写一个函数来查找字符串的长度，因为不允许包含头文件）

int mystrlen(const unsigned char *string)
{
  int length = 0; /* holds the value of the length of the string */

   /* goes through elements of string not counting the null character */
  while(*string != '\0')
  {
    string++;  /* moves to the next letter of the string */
    length ++; /* counts how many letters there are      */
  }
  return length; /* returns the length of the string */
}

void jumble(unsigned char *string, const unsigned char *password, 
            enum CODE_METHOD method, int passes)
{
  if(method == ENCODE)
  {
    int i;
    for(i = 1; i <= passes; i++)
    {
      while(*string != '\0')
      {
        *string = *string + *password;

        string++;
        password++;
      }
    }
  }
  else
  {
    int i;
    for(i = 1; i <= passes; i++)
    {
      while(*string != '\0')
      {
        *string = (*string) - (*password);

        string++;
        password++;
      }
    }
  }
}

这是正确的输出：

Test0 ======================================
length of a is 1
length of b is 4
length of c is 0
length of d is 174

Test1 ======================================
Original phrase:
THIS IS A SECRET.

Encoded phrase:
xkstDl}AeC}fguouR

Test2 ======================================
Encoded phrase:
xkstDl}AeC}fguouR

Decoded back:
THIS IS A SECRET.

这就是我得到的：

Test0 ======================================
length of a is 1
length of b is 4
length of c is 0
length of d is 174

Test1 ======================================
Original phrase:
THIS IS A SECRET.

Encoded phrase:
xkst ┴╛ô╡@╓àh1

Test2 ======================================
Encoded phrase:
xkstDl}AeC}fguouR

Decoded back:
THISD5° e1ocp!

再次感谢您！

Answer 1

由于您的文本字符串比密码字符串长，因此您的密码字符串末尾超出了限制（例如，文本为8个字符，但您只有4个密码字符）。这是未定义的行为（例如，此后为随机字符）

以下是一些代码，当密码到达末尾时，这些代码会返回密码的开头。我只是猜测这就是算法所需要的。您可能需要对其进行调整，以匹配所需的确切算法/结果。

此外，每次通过都必须将string变量重置为开头。否则，只有 first 通过才会生效。

void
jumble(unsigned char *string, const unsigned char *password, enum CODE_METHOD method, int passes)
{
    int pwlen;
    unsigned char *str;
    unsigned char *pw;

    pw = password;

    if (method == ENCODE) {
        int i;

        for (i = 1; i <= passes; i++) {
            //pw = password; ?
            str = string;
            while (*str != '\0') {
                *str = *str + *pw;
                str++;
                if (*++pw == 0)
                    pw = password;
            }
        }
    }
    else {
        int i;

        for (i = 1; i <= passes; i++) {
            //pw = password; ?
            str = string;
            while (*str != '\0') {
                *str = (*str) - *pw;
                str++;
                if (*++pw == 0)
                    pw = password;
            }
        }
    }
}

C语言中字符串的加减法（编码/解码）

1 个答案: