Question

试图进行凯撒密码。

enum alfabeto{
    A=0,B,C,D,E,F,G,H,I,L,M,N,O,P,Q,R,S,T,U,V,Z         // 21
};

void cifra(char *string, int k){
    enum alfabeto letter;    // initialize the enum letter
    size_t i = 0;    // initialize counter
    while (*(string+i)!='\0'){    // while string is not ended
        letter = *(string+i);     // attempt to "link" the enum letter to the equivalent (already uppercased) char
        printf("%d", letter);
        letter = (letter+k) % 21;    // then it increases of a factor k and if it goes out of 21, it should take the right value
        printf(" %d\n", letter);
        ++i;
    }
}

输出：

$ ./"cesare" 

write the text:
>TEST

choose the factor k:
>2

84 8
69 14
83 7
84 8

值是错误的...可能是因为我无法将枚举值“链接”到字符...我该怎么办？c

Answer 1

    letter = *(string+i);     // attempt to "link" the enum letter to the equivalent (already uppercased) char

应该是：

    letter = *(string+i) - 'A';     // attempt to "link" the enum letter to the equivalent (already uppercased) char

这样，“ A”将根据需要映射为零。

Answer 2

//                         A B C D E F G H I  J  K L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z
const int offsetTable[] = {0,1,2,3,4,5,6,7,8,-1,-1,9,10,11,12,13,14,15,16,17,18,19,-1,-1,-1,20};
const char charLookupTable[] = "ABCDEFGHILMNOPQRSTUVZ";

// Should probably use a struct for the above to prevent size mis-matches
const int offsetTableSize = 21;

void cifra(char *str, int k){
  char letter;
  int newLetterOffset;

  printf("%s\n", str);  // Show initial string.

  while (letter = *str++){    // while string is not ended
    const int oldLetterOffset = offsetTable[letter - 'A']; // Get the old letter's offset.
    if(letter <= 'A'  || letter >= 'Z' || oldLetterOffset == -1)
    {
        printf("\nBad character %c found - ending string processing\n", letter);
        return;
    }
    newLetterOffset = (oldLetterOffset + k) % offsetTableSize; // Get the letter's position, add the global offset, and wrap to table size.
    printf("%c", charLookupTable[newLetterOffset]);  // Use that offset to read the character's value from the shortened alphabet.
  }
  printf("\n\n");
}

int main()
{
  cifra("HELLO", 0);
  cifra("HELLO", 1);
  cifra("HELLo", 1);
  cifra("YELLO", 1);

  return 0;

}

注意，我需要2个表来完成这项工作，因为我们不得不进出缩短的字符集。通常，我们将使用一个结构来保存这两个数组，但是在此示例中，我将其保持简单。另外，数组不必是全局的，但我也将它们放在那里以使事情更简单。

注意，我还更改了您的printf（）值，以使用字符和字符串以使其更易于阅读。

最后，我添加了一些错误检查，因为您不能信任用户为您提供良好的字符串。由此导致许多安全漏洞或随机崩溃。

将字母与枚举进行比较

2 个答案: