为什么在使用凯撒密码加密时不能打印 space?

Why can't I print space while encrypting using Caesar cipher?

我正在尝试编写一个使用凯撒密码方法加密或解密消息的 C 程序。用户可以输入带有空格的消息,但我的 C 程序会打印一些其他字符(例如 []、字母符号或有时是字母表)。谁能建议更改我的代码以打印空格?

#include <stdio.h>
#include <conio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>

void main()
{
    int  i,j,s,k,p,choice,key,n,count;
    char alpha[27]={'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'};
    char msg[100],encrypt[100];
    printf("\t\t\tCEASER CIPHER");
    printf("\n\t\t\t-------------");
    printf("\n1.Encrypt");
    printf("\n2.Decrypt");
    printf("\nEnter the choice:");
    scanf("%d",&choice);
    switch(choice){
    case 1 :{system("cls");
             printf("Enter the message to be encrypted:");
             fflush(stdin);
             gets(msg);

             printf("\nEnter the shift key:");
             scanf("%d",&key);
             n=strlen(msg);
             printf("\nThe encrypted key is :");
            for(i=0;i<n;i++){
                    count=0;
                for(j=0;j<27;j++){
                if(msg[i]==alpha[j]){
                s=j+key;
                k=s%26;
                encrypt[i]=alpha[k];
                count=1;
                break;
                }
             }
              if(count=1)
                printf("%c",encrypt[i]);
              else if(count=0)
              printf("-");
             }
             }
    case 2 :{
            }
             }
}
1.output:-
Enter the message to be encrypted:hello world
Enter the shift key:3
Th encrypted key is:khoor'zruog
expected output:-
Enter the message to be encrypted:hello world
Enter the shift key:3
Th encrypted key is:khoor-zruog
1.output:-
Enter the message to be encrypted:how are you
Enter the shift key:3
Th encrypted key is:krz-duhabrx
expected output:-
Enter the message to be encrypted:how are you
Enter the shift key:3
Th encrypted key is:krz-duh-brx

你做的基本上是正确的,但不是正确的做法。您正在使用两个循环来检查字母是否在上面的数组中。您可以像这样轻松检查:

if (msg[i] >= 'a' && msg[i] <= 'z')  // if msg[i] is letter

所以,像这样编写代码会更容易:

    for (int i = 0; i < n; ++i)
    {
        if (msg[i] >= 'a' && msg[i] <= 'z')
        {
            encrypt[i] = msg[i] + key;

            if (encrypt[i] > 'z')
                encrypt[i] = 'a' + (encrypt[i] - 'z');  // or encrypt[i] -= 26;
        }
    }

而且,正如 Johathan 提到的,您需要 breakcases 之间。此外,正如他再次所说,if (count = 1) 不检查 count 的值:它只是分配它。如果您需要比较,请使用 ==(尽管您并不需要它)。