Incrementing Char Type In Java

This is the equivalent program of your program:

public class Tester {
    public static void main(String args[]){
             char start='\u0041';
             char next='\u0041'+1;
             System.out.println(next);
    }
}

But as you see, next=start+1, will not work. That's the way java handles.

The reason could be that we may accidentally use start, with integer 1 thinking that start is an int variables and use that expression. Since, java is so strict about minimizing logical errors. They designed it that way I think.

But, when you do, char next='\u0041'+1; it's clear that '\u0041' is a character and 1 will be implicitly converted into a 2 byte. This no mistake could be done by a programmer. May be that's the reason they have allowed it.

char is 2 bytes in java. char supports unicode characters. When you add or subtract a char var with an offset integer, the equivalent unicode character in the unicode table will result. Since, B is next to A, you got B.

In Java, char is a numeric type. When you add 1 to a char, you get to the next unicode code point. In case of 'A', the next code point is 'B':

char x='A';
x+=1;
System.out.println(x);

Note that you cannot use x=x+1 because it causes an implicit narrowing conversion. You need to use either x++ or x+=1 instead.

A char is in fact mapped to an int, look at the Ascii Table.

For example: a capital A corresponds to the decimal number 65. When you are adding 1 to that char, you basicly increment the decimal number by 1. So the number becomes 66, which corresponds to the capital B.