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Which of the following results in a negative value of a? A: int a = -1; a = a >>> 8; B: byte a = -1; a = (byte)(a >>> 5); C: int a = -1; a = a >> 5; D: int a = -1; a = a >>> 32

The answer is supposed to be A, B and D. I don't see this at all. I would think the only negative one would B. I believe that A, C and D are incorrect because any time an int is unsigend-right-shifted, a positive number results. In the case of B above though things are sort of different. As I recall when a non-int is acted on by any shift operation, it is first cast to an int and then the shifting is performed. This converts the byte of -1 (11111111) into the int form of -1 (11111111111111111111111111111111). This is unsigned-right-shifted by 5, which results in some positive number (00000111111111111111111111111111). Which is then cast back into the form of a byte which leaves us with -1 again (11111111 that is). Am I right here?

I wrote the following program: public class Test{ public static void main(String[] args){ int A = -1; A = A >>> 8;

byte B = -1; B = (byte)(B >>> 5);

int C = -1; C = C >> 5;

int D = -1; D = D >>> 32;

System.out.println("A = " + A); System.out.println("B = " + B); System.out.println("C = " + C); System.out.println("D = " + D); } } And got these results: A = 16777215 B = -1 C = -1 D = -1

Honestly, this makes no sense to me. I would think that C is the only right answer, since >>> is the unsigned right shift (it fills the bits to the left with 0) and >> fills the left with whatever the sign bit is. In Two's Complement, negative numbers have a 1 as their Most Significant Bit. Since each of these start as negative (with MSB of 1), only choice C should stay negative, since it will be the only one left with a MSB of 1 (because A, C, D fill with 0). Can someone please explain this? I'm thoroughly confused.

Just taking a quick look without getting out my pencil and paper... B *could* be correct since you're doing a cast after the shift and just grabbing the rightmost 8 bits - the 8th bit might be a 1 C is using the signed shift operator - so the sign bit will be replicated D - The right hand shift operator can only be, in practice, from 0-31 for ints and from 0-63 for longs. Think about it like using the remainder operator - in this case the right operand (32) becomes 0, so no shifting occurs! good luck in your studies!

Spot false dilemmas now, ask me how!
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The signed bit is the left most bit so >>> changes a 1 to 0 but not from 0 to 1, the >> keeps the sign bit so if it was 1 all bit shifted will be 1's and 0 with 0's, hopefully the bold has worked below

Which of the following results in a negative value of a? A: int a = -1; a = a >>> 8; B: byte a = -1; a = (byte)(a >>> 5); C: int a = -1; a = a >> 5; D: int a = -1; a = a >>> 32

A is: 11111111 11111111 11111111 11111111 >>> 8 gives you 00000000 11111111 11111111 11111111 this results in a big poistive int number. B is: converts the byte to an int then does the shifting, then casts back to a byte so bits look like this: byte 11111111 promoted to 11111111 11111111 11111111 11111111 then shift to get 00000111 11111111 11111111 11111111 but we convert back to get the last 8 bits on the right hand side which is 11111111 byte -1 C is: 11111111 11111111 11111111 11111111 then shift with which ever the left most bit was which is 1 to get the same result D is: 11111111 11111111 11111111 11111111 by 32 bits, I needed the pencil first time round, but then i noticed that it can only shift 31 times to give you +1 but 32 times is shifting 0 times so hence the same result of -1. I normally find it easier writing out the bits first then trace out what will happen to the bits. I hope this helps guys Davy [ February 18, 2004: Message edited by: Davy Kelly ]