java puzzlers from oca part 4

In the fourth part of Java Puzzlers, we have something related to char type.

public class Puzzler {

    public static void main(String[] args){
        char myChar = 'a';
        myChar++;

        System.out.println(myChar);
    }
}

You may have guessed it. It will print "b" and the reason for it is that char type is unsigned numeric primitive in the disguise of a character. So if I add one then I'll get the next character in unicode representation.

Then let's take a look at that one

public class Puzzler {

    public static void main(String[] args){
        char myChar = 'a';

        System.out.println(myChar + myChar);
    }
}

Will this print "aa"? Or  which's 97 + 97 = 194 (where 97 is value of 'a'). I don't know if you guessed it right but the result is neither. It's "194". When Java sees plus it tells "hmm that's an addition not a concat" and adds myChars up and returns the int value for it.

java puzzlers from oca part 3

In this third part of Java puzzlers, we will see a surprise in variable naming restrictions.


If I show you this, I'm sure you won't be surprised that this does not compile. static is one of the reserved keywords so why should it work?

public class Puzzler {

    public static void main(String[] args){

        int static = 2;
    }

}

Now I'll ask you a more difficult one. What you think about the below code. Will this compile?

public class Puzzler {

    public static void main(String[] args){
        int bool = 0;
        int integer = 1;
        int const = 2;
        int goto = 3;
    }
}

None of these should be reserved keyword. This is not C right? If you thought that it will compile, you're wrong. const and goto are reserved keywords, but bool and integer are fine.