bitwise OR on short and long (java)

Asked Apr 28 '16 at 21:50

Active Apr 28 '16 at 22:10

Viewed 34 times

The following code calculates the parity of number.

The question is what happens in the line count^=(number&1). The bitwise AND operation will result in a long. The a bitwise OR is performed between a short (count) and a long (number&1), which should result in a long. Why can the result still be saved in a short valiable (count)?

long number=17;

public short parityCalc() {
    short count=0;

    for(int c=0;c<64;c++){
        count^=(number&1);
        number>>>=1;
    }
    return count;
}

edited Apr 28 '16 at 22:10

Sotirios Delimanolis

274,122
60
696
724

asked Apr 28 '16 at 21:50

Hargen.Hans

The question had nothing to do with the compound assignment operator. I changed the question now so that the compound assignment operator is left out. – Hargen.Hans Apr 28 '16 at 21:56
Your new code doesn't compile. It only compiled because of the compound assignment operator. Read the duplicate. – Sotirios Delimanolis Apr 28 '16 at 21:57
what is "number>>>=1;" ? – 2ARSJcdocuyVu7LfjUnB Apr 28 '16 at 21:59
To be clear, I'm referring to the compound `^=` operator in the first revision of your question. – Sotirios Delimanolis Apr 28 '16 at 22:01
Thanks, Sotirios! It was indeed a duplicate. It turns out that the compound assignment operator is also performing a cast. – Hargen.Hans Apr 28 '16 at 22:02

bitwise OR on short and long (java)

0 Answers0