方法在java中返回int而不是char

Question

Like stated in the title I have a problem with a method in java that I wrote. This is the code:

public static char shift(char c, int k) {

    int x = c;

    int d = c - 65 + k;
    int e = c - 97 + k;

    if (x > 64 && x < 91 && d >= 0 ) {

        c = (char) ( d % 26 + 65);

    } else if (x > 96 && x < 123 && e >= 0 ) {

        c = (char) (e % 26 + 97 );
    }


    if (x > 64 && x < 91 && d < 0 ) {

        c = (char) ( (d + 26) % 26 + 65);

    } else if (x > 96 && x < 123 && e < 0 ) {

        c = (char) ( (e + 26) % 26 + 97);
    }

    return c;
}

I want to shift a letter in Alphabet. The code works perfectly if I use it like this (Caesar Chiper):

String s = " ";
    String text = readString();
    int k = read();

    for (int i = 0; i < text.length(); i++) {

        char a = text.charAt(i);
        int c = a;

        int d = a - 65 + k;
        int e = a - 97 + k;

        if (c > 64 && c < 91 && d >= 0 ) {

            a = (char) ( d % 26 + 65);

        } else if (c > 96 && c < 123 && e >= 0 ) {

            a = (char) (e % 26 + 97 );
        }
          if (c > 64 && c < 91 && d < 0 ) {

            a = (char) ( (d + 26) % 26 + 65);

        } else if (c > 96 && c < 123 && e < 0 ) {

            a = (char) ((e + 26) % 26 + 97);
        }

        s += a;
    }

    System.out.println(s);
}

I don't understand why the method shift returns an integer when I use it like this: shift('c', 5); it returns 104, which is the dec number for h. I'm a beginner in java and a slow person.

Thank you in advance.

Answer 1

Your mistake is that you're probably adding the unicodes of two characters. This can happen if you do something like this:

System.out.println('b' + 'a');

or in your case

System.out.println(shift('c', 5) + 'a');

---

To get the desired result, convert the char to a string before printing:

String result = Character.toString(shift('c', 5));
System.out.println(result + 'a');

or

System.out.println(shift('c', 5) + "a");