Hi, What will be the result of 1<<32....and why is it equal to 1>>32 or 1>>>32. As per my understanding, the bits will be shifted left and after 31 shifts the number will be reduced to zero as then it will be filled with zeros from right. Am i right? In what conditions can unsigned right shift operator produce negative values?

1 << 32 The left shift (<< shifts all bits to the left, filling the right side with zeroes, 1>>32 right shift (>> shifts all bits right, filling in the left side with whatever the sign bit was. OR 1>>>32 The unsigned right shift (>>> moves all bits to the right, but fills the left side with zeroes, Regards PREM

Hi, I think shifting the bits by 32 places, either left or right will result in the same number. Since, shift by number greater than or equal to 32 will always be a modulo of 32. That is, 1>>32 will be 1>>0 (1 >> (32%32)) , resulting in 1 itself. Is my assumption right?? Thanks, Uma...

Hi Uma, you are right but you didnt give him the complete picture. Hi Harjinder, in case of integers, it works like this: opr1>>opr2 where opr2 = (opr2%32) if opr2 >= 32. for both left shift and right shift. in case of long, it works like this: opr1>>opr2 where opr2 = (opr2%64) if opr2 >= 64. for both left shift andirght shift. now just think about short and try out how it works for a short?

Originally posted by Arun Subbu: now just think about short and try out how it works for a short? [/QB]

byte, short, char is the same as int i.e. opr2=opr2%32 A unary numeric promotion to int is performed for these primitive types when used in a shift operation.

Don't get me started about those stupid light bulbs.