CODEWITHSUNDEEP
java
Ullas
Ullas

Reputation: 837

Regarding shift operators

public class Shift{

    public static void shift1(){
        int i = 0;
        while(-1 << i != 0){
            i++
        }
    }

    public void shift2(){
        for(int i=-1;i!=0;i<<=1){
            System.out.println(i);
        }
    }
}

The first method goes to infinite loop where as second iterates 31 times to become 0;why does java only consider the lower order 5 bits of right operand while shifting?

Upvotes: 2

Views: 97

Answers (3)

Bohemian
Bohemian

Reputation: 424983

The "lowest 5 bits" is not relevant.

-1 is represented as 32 "1" bits: 11111111111111111111111111111111

Each << 1 operation chews up one of those bits until i is zero - that's what's controlling the number of iterations.

Upvotes: 0

qaphla
qaphla

Reputation: 4733

Shifting by 32 or more isn't done by Java both because it's a fairly useless operation on ints, and simply because that is how it is specified to work.

Upvotes: 2

user207421
user207421

Reputation: 310859

Because that's what is specified in the Java Language Specification: "If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance."

A lot of hardware works that way as well.

Upvotes: 5

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