Question

我正在为C中的一个类加密程序。我使用modulo 27数学来执行加密。我将每个加密的字符添加到一个数组中，但我注意到未打印的字符也会在最后添加到我的字符串中。当我检查加密文本的字数时，我意识到这一点，它包含的字符多于加密的原始文本。任何人都可以解释为什么会这样吗？这考虑了文本末尾的换行符。

Plaintext = THE RED GOOSE FLIES AT MIDNIGHT STOP  - wc is 37
Ciphertext = ACBVKWNOGMMMPQHNI XL QBJXDPNVIQVSNZN - wc is 40

   //Go through each character of the plaintext
    for (i = 0; i < size; i++)
    {
      //Convert the characters to an integer
       PlainNums[i] =  charInt(plaintext[i]); 
       KeyNums[i] = charInt(key[i]); 

    //Add the int from plain text and the key together
    CipherNums[i] = PlainNums[i] + KeyNums[i];
    if (CipherNums[i] > 27) //Wrap around if the number exceeds 27
    {
        CipherNums[i] -= 27;
    }
    CipherNums[i] = CipherNums[i] % 27; //Use modulo 27 math to generate a new integer

    cipherText[i] = IntChar(CipherNums[i]);
}

  int charInt(char c)
  {
   switch (c) 
   {
    case 'A': return 0;
    case 'B': return 1;
    case 'C': return 2;
    case 'D': return 3;
    case 'E': return 4;
    case 'F': return 5;
    case 'G': return 6;
    case 'H': return 7;
    case 'I': return 8;
    case 'J': return 9;
    case 'K': return 10;
    case 'L': return 11;
    case 'M': return 12;
    case 'N': return 13;
    case 'O': return 14;
    case 'P': return 15;
    case 'Q': return 16;
    case 'R': return 17;
    case 'S': return 18;
    case 'T': return 19;
    case 'U': return 20;
    case 'V': return 21;
    case 'W': return 22;
    case 'X': return 23;
    case 'Y': return 24;
    case 'Z': return 25;
    case ' ': return 26;
    default:  return -1;
   }
}

char IntChar(int n)
{
  switch(n)
  {
    case 0: return 'A';
    case 1: return 'B';
    case 2: return 'C';
    case 3: return 'D';
    case 4: return 'E';
    case 5: return 'F';
    case 6: return 'G';
    case 7: return 'H';
    case 8: return 'I';
    case 9: return 'J';
    case 10: return 'K';
    case 11: return 'L';
    case 12: return 'M';
    case 13: return 'N';
    case 14: return 'O';
    case 15: return 'P';
    case 16: return 'Q';
    case 17: return 'R';
    case 18: return 'S';
    case 19: return 'T';
    case 20: return 'U';
    case 21: return 'V';
    case 22: return 'W';
    case 23: return 'X';
    case 24: return 'Y';
    case 25: return 'Z';
    case 26: return ' ';
    default:  return '!'; //error
 }

}

Answer 1

尝试在循环结束时添加cipherText[size] = '\0';以确保编码的字符串仅包含编码数据。

Answer 2

使用代码中的cipherText在for loop之前初始化memset，如下所示。

memset(cipherText,0,sizeof(cipherText)); 

 for (i = 0; i < size; i++)
    {
      //Convert the characters to an integer
       PlainNums[i] =  charInt(plaintext[i]);
      //....
     // rest of the code

您必须include<string.h。它有效。

附加到字符串的未打印字符

2 个答案: