Converting ASCII code to char in Java

Question

Here's my code below and it prints J=74, A =65, and M=77. How do I get it to print just the characters K, B, N as the result of moving down the alphabet?

BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
String string = JOptionPane.showInputDialog( " Please Enter Code " );

for (int i = 0; i < string.length (); ++i) {
    char c = string.charAt(i);
    int j = (int)c;
}


System.out.println("ASCII OF "+c +" = " + j + ".");

Answer 1

Simply casting int to char

System.out.println((char) 65);        // A
System.out.println((char) ('A' + 1)); // B
System.out.println((int) 'A');        // 65

Beware, this is a raw attempt at a naive problem (or at least a bad spoken one). The last line of your snippet already contains everything you need. Maybe you only miss that char in Java is really an integer type, so you can use char literals with operators like + or even %

System.out.println((char) ('Z' + 5));
System.out.println((char) ('Z' / 2));
System.out.println((char) ('Z' % 31));

Answer 2

I think you have already given ur answer in question itself. You should cast the integer to char as below

    int j=77;
    char c=(char)j;
    System.out.println(c);