Java Char increment

Question

I was executing the following code

class First
{

 public static void main(String arg[])
        {
            char x= 'A';
            x = x++;
            System.out.println(x);
        }
}

Here the output is A. My question is why didn't x get incremented before printing.

What if your line `x = x++;` just said `x++;`? What happens then? — Carl Manaster, Nov 19 '14 at 17:04
It did. It didn't get incremented before *assigning*, though. — Dave Newton, Nov 19 '14 at 17:06

score 3 · Accepted Answer · answered Nov 19 '14 at 17:04

You're using the post-increment operator incorrectly - you don't need to use assignment as well. And in this case it undermines what you're trying to do.

For context, remember that the post-increment operator increases the value, and returns the old value. That is, x++ is roughly equivalent to:

int x_initial = x;
x = x + 1;
return x_initial;

Hopefully now you can see why your code is failing to change x. If you expand it, it looks like:

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

and the net effect of the assignment, is to set x back to what it was originally.

To fix - you can simply call x++. Or, if you want to make it clear there's some sort of assignment happening, x += 1 or even just x = x + 1 would do the same thing.

score 2 · Answer 2 · answered Nov 19 '14 at 17:03

class First
{

 public static void main(String arg[])
        {
            char x= 'A';
            x = x++;   // it is post increment as ++ sign is after x
            System.out.println(x);
        }
}

Post Increment(x++) : First execute the statement then increase the value by one.

Pre Increment (++x) : First increase the value by one then execute the statement.

Java Char increment

2 Answers2