我如何将这些字符沿字母表的 3 个空格转移而不进入不属于字母表的其他字符？

Question

因此，我必须将收到的每个字符沿字母表向下移动 3 个空格，当它到达字母表的末尾时，它会返回。所以 A 就是 X，B 就是 Y……等等。我唯一的问题是当它经过 z 时，它会打印 ^ _ ' 知道为什么会这样吗？我知道这与我的 encrypt_file 函数中的循环变量有关。是的，我知道该功能不完整，因为它不寻找大写字母，我只是想把我的测试用例记下来，这样我几乎可以复制并粘贴它的其余部分。

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <strings.h>

#define OP_ENCRYPT "-e"
#define OP_DECRYPT "-d"

enum method
{
 ENCRYPT,
 DECRYPT
};

typedef int (*converter_t)(int);

int encrypt_file(int c)
{
 if(c >= 'a' && c <= 'z') 
 {
  if((c - 3) >= 'z')
  {
   return c + 3;
  }
  else
  {
   return c - 3;
  }
 
 }
 else
 {
  return c;
 } 
}


int decrypt_file(int c)
{
 int j; 
 if(c<= 'A' && c>= 'Z')
 {
  if((j = c - 3) >= 'Z')
  { 
   c = j; 
   return c; 
  } 
  else
  {
   j = c + 3;
   c = j; 
   return c;
  }
 }
 else if(c<= 'a' && c>= 'z')
 {
  if((j = c - 3) >= 'z')
  { 
   c = j; 
   return c;
  } 
  else
  {
   j = c + 3;
   c = j; 
   return c;
  }
 }
 return c; 
}


enum method conversion_method(const char *arg)
{
 if (strcmp(arg,OP_ENCRYPT) == 0)
 {
  return ENCRYPT;
 }
 if (strcmp(arg,OP_DECRYPT) == 0)
 {
  return DECRYPT;
 } 
 else
 {
  return ENCRYPT;
 }
}

converter_t converter(enum method mode)
{
 switch (mode)
 {
  case ENCRYPT:
   return encrypt_file;
  case DECRYPT:
   return decrypt_file;
  default:
   return encrypt_file;
 }
}
int is_option(const char *str)
{
 return (strcmp(str, OP_ENCRYPT) == 0) || (strcmp(str,OP_DECRYPT) == 0); 
}

void convert( FILE *src, FILE *dest, converter_t method)
{
 unsigned int ch;

 while ((ch = fgetc(src)) != EOF)
 {
  fprintf(dest, "%c", (*method)(ch));
 }
}

int main(int argc, const char *argv[])
{
 FILE *src = stdin;
 FILE *dest = stdout;
 enum method method = ENCRYPT;
  
 if(argc ==2 ) 
 {
  if (is_option(argv[1])) 
  {
   method = conversion_method(argv[1]);
  }
  else
  {
   src = fopen(argv[1], "r");
   if (src == NULL)
   {
    fprintf(stderr, "%s: unable to open %s\n", argv[0], argv[1]);
    exit(EXIT_FAILURE);
   }
  }
 }
 else if (argc == 3) 
 {
  if (is_option(argv[1]))
  {
   method = conversion_method(argv[1]);
   src = fopen(argv[2], "r");
   if (src == NULL) 
   {
    fprintf(stderr, "%s: unable to open %s\n", argv[0], argv[2]);
    exit(EXIT_FAILURE);
   }
  }
  else
  {
   fprintf(stderr, "%s: invalid option %s\n", argv[0], argv[1]);
   exit(EXIT_FAILURE);
  }
 }

 convert(src, dest, converter(method));

 return EXIT_SUCCESS;
}

Answer 1

你想得 waaay 比你需要的更难，你迷路了。在这里，检查一下：

#include <stdio.h>
#include <assert.h>

int encrypt_char( int c )
{
    assert( c >= 'a' && c <= 'z' );
    c -= 3;
    if( c < 'a' )
        c += 26;
    return c;
}

int decrypt_char( int c )
{
    assert( c >= 'a' && c <= 'z' );
    c += 3;
    if( c > 'z' )
        c -= 26;
    return c;   
}

void encrypt_text( char* ptext )
{
    for( ; *ptext != 0; ptext++ )
        *ptext = encrypt_char( *ptext );
}

void decrypt_text( char* ptext )
{
    for( ; *ptext != 0; ptext++ )
        *ptext = decrypt_char( *ptext );
}

static char text[256] = "abcdefghijklmnopqrstuvwxyz";

int main( int argc, const char *argv[] )
{
    printf( "ORIGINAL:  %s\n", text );
    encrypt_text( text );
    printf( "ENCRYPTED: %s\n", text );
    decrypt_text( text );
    printf( "DECRYPTED: %s\n", text );
    return 0;
}

我如何将这些字符沿字母表的 3 个空格转移而不进入不属于字母表的其他字符？

How would I transfer these characters down the alphabet 3 spaces without going into other characters that aren't apart of the Alphabet?

c

encryption

non-ascii-characters