不能将凯撒密码移动超过 1

Question

我是 Java 的新手，我必须编写一个凯撒密码程序。这是我的代码。我遇到的问题是我不能将它移动超过 1。例如，如果我写 'help me' 并输入 2 的移动值，它仍然只移动 1 --> 'ifmq nf'

  public static void main(String[] args) 
    {
        Scanner sc=new Scanner(System.in);
        String str;
        String key;
        int keyLength;

        System.out.println("Enter message:");
        str=sc.nextLine();
        System.out.println("Enter encryption key:");
        key=sc.next();
        keyLength=key.length();
        //This for loop is repeated use of 'Enrypt' and 'Decrypt' options
        for(;;)
        {
            System.out.println("1.Encrypt\n2.Decrypt\n3.Exit...");
            int choice=sc.nextInt();
            switch(choice)
            {
                case 1:
                /*send input string keyLength to encrypt() method to encrypt it returns 'Encrypted' string*/
                    System.out.println("Encrypted message..."+encrypt(str,keyLength));
                    break;
                case 2:
                    //send retrived string from encrypt() method and keyLength to decrypt() method it returns 'Decrypted' string
                    System.out.println("Decryptedmessage..."+decrypt(encrypt(str,keyLength),keyLength));
                    break;
                case 3:
                    //exit from the program
                    System.exit(0);
                    break;
                default:
                System.out.println("Invalid option..");
            }
        }
    }
    public static String encrypt(String str,int keyLength)
    {
        String encrypted="";
        for(int i=0;i<str.length();i++)
        {
            //stores ascii value of character in the string at index 'i'
            int c=str.charAt(i);
            //encryption logic for uppercase letters
            if(Character.isUpperCase(c))
            {
                c=c+(keyLength%26);
                //if c value exceeds the ascii value of 'Z' reduce it by subtracting 26(no.of alphabets) to keep in boundaries of ascii values of 'A' and 'Z'
                if(c>'Z')
                    c=c-26;
            }
            //encryption logic for lowercase letters
            else if(Character.isLowerCase(c))
            {
                c=c+(keyLength%26);
                //if c value exceeds the ascii value of 'z' reduce it by subtracting 26(no.of alphabets) to keep in boundaries of ascii values of 'a' and 'z'
                if(c>'z')
                    c=c-26;
            }
            //concatinate the encrypted characters/strings
            encrypted=encrypted+(char) c;
        }
        return encrypted;
    }
    public static String decrypt(String str,int keyLength)
    {
        String decrypted="";
        for(int i=0;i<str.length();i++)
        {
            //stores ascii value of character in the string at index 'i'
            int c=str.charAt(i);
            //decryption logic for uppercase letters
            if(Character.isUpperCase(c))
            {
                c=c-(keyLength%26);
                //if c value deceed the ascii value of 'A' increase it by adding 26(no.of alphabets) to keep in boundaries of ascii values of 'A' and 'Z'
                if(c<'A')
                    c=c+26;
            }
            //decryption logic for uppercase letters
            else if(Character.isLowerCase(c))
            {
                c=c-(keyLength%26);
                //if c value deceed the ascii value of 'A' increase it by adding 26(no.of alphabets) to keep in boundaries of ascii values of 'A' and 'Z'
                if(c<'a')
                    c=c+26;
            }
            //concatinate the decrypted characters/strings
            decrypted=decrypted+(char) c;
        }
        return decrypted;
    }

Answer 1

正如 Scary Wombat 和 MadProgrammer 所说，在第 12 行你有 keyLength=key.length();。这个returns你输入的String的length。所以如果你输入2，keyLength就是1。如果你输入11， keyLength 是 2.

我会完全摆脱 keyLength，将 key 设为 int 而不是 String，然后设置 key=sc.nextInt();

这将允许 1 为 1，2 为 2，依此类推。

记得将 keyLength 的所有实例更改为 key

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