CODEWITHSUNDEEP
java
Dario
Dario

Reputation: 147

Why does this code print 100 instead of 1?

public class Short {
    public static void main(String[] args) {

        Set s = new HashSet();

        for(short i = 0; i < 100; i++) {
            s.add(i);
            s.remove(i-1);
        }

        System.out.print(s.size());
    }

}

Can anyone tell me why it prints 100 instead of 1?

Upvotes: 4

Views: 199

Answers (2)

MadProgrammer
MadProgrammer

Reputation: 347234

There seems to be some auto boxing going on...that is Java is automatically converting between Object and primitive...

If I ... rename your class, use Short instead of short in the initialisation of the Set and then use...

Set<Short> s = new HashSet<Short>();

for (short i = 0; i < 100; i++) {
    s.add(i);
    s.remove(i - 1);
}

System.out.println(s.size());

It will print 100...but why?

To answer that, we need to take a closer look at the remove method...

Set#remove(Object o) expects an Object, not the generic type like add, but an actual Object...when you do i - 1, Java assumes that 1 is an int and automatically scales the types up and auto boxes it as new Integer(i - 1)...which clear does not exist within the set (you don't have any Integer objects!)

However, if we change s.remove(i - 1); to s.remove((short)(i - 1));, we force the conversion of the value back to short which then gets autoboxed as new Short(i - 1), which does exist in your set and the end result is it will now print 1...

Simple ;)

Upvotes: 10

bcsb1001
bcsb1001

Reputation: 2907

Upon running this code, I found that, after converting your primitive generic to java.lang.Short, the problem is when you do i-1. short-int returns int, therefore the remove operation tries to remove an Integer from s. int and short, and, respectively, Integer and Short are very different.

Upvotes: 2

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