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Because >> operator propagates the sign, while >>> adds zeros to the most significant bits whatever may be the sign. Ankur

Daniel Liu
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Joined: Jul 05, 2000
Posts: 19

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Would you please explain more what the word "propagate" means?

Originally posted by Ankur: Because >> operator propagates the sign, while >>> adds zeros to the most significant bits whatever may be the sign. Ankur

Ankur Gupta
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Joined: Jun 13, 2000
Posts: 66

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It means that >> operator will put the same sign in the most significant bits i.e. for +ve nos. it will put zeros and for -ve nos it will put 1 in the most significant bits.

Daniel Liu
Greenhorn

Joined: Jul 05, 2000
Posts: 19

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Thanks. That is what i do't know.

Originally posted by Ankur: It means that >> operator will put the same sign in the most significant bits i.e. for +ve nos. it will put zeros and for -ve nos it will put 1 in the most significant bits.

I still don't understand. -1 is 11111111111111111111111111111111 (32 1's)in binary. Shifting one position to the left gives 1111111111111111111111111111111 (31 1's). How can the value stay at -1?

Please check the following code: class Test { public static void main(String[] args) { int a=-1; System.out.println(Integer.toBinaryString(a)); a=a>>1; System.out.println(Integer.toBinaryString(a)); a=a>>>1; System.out.println(Integer.toBinaryString(a));

} } OUTPUT: a)32 1's(for a=-1) b)32 1's(for a=a>>1) c)31 1's(for a=a>>>1) At b,though the right side is shifted by one bit,the left hand side is always filled with the value of the most significant bit (here in this case it is one for -1 in binary).Whereas at c,the left side is filled with zero and not by the value of most significant bit(which is still one ).