Converting ASCII code to char in Java

java

Solution

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));

Problem

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 + "."); ```

Original source