I tried this following code class Q2{ public static void main(String args[]) { int a = 5; int b = -2; int c=a>>b; int d=a<<b; int e=a>>>b; System.out.println(+c); System.out.println(+d); System.out.println(+e); } } After compilation and running I am getting the output as, 0 1073741824 0 Pls explain me the answer.

Try this:- System.out.println( (5 >> -2) == (5 >> 30) ); JLS 15.19 Shift Operators 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. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (�15.22.1) with the mask value 0x1f.

Hi Ramki: Try these simple rules that cover most of the sitution. Please let me know if find they do not work in any situation: Shift Operators A op B Assuming A is any integral value except long. For long A, replace 32 with 64: 1. B = B%32 2. If B is negative and not –32 : B=B+32 3. For signed right shift operator (>> , divide A successively by 2, B number of times. Round down the final result if A is positive. Round up the final result if A is negative. 4. For signed left shift operator (<< ; multiply A succesively by 2, B number of times. If (32-B)th bit of original A value is 1, make the result –ve if it is not already negative. If the final result is more than Integer.MAX_VALUE, make final result 0. 5. For unsigned right shift operator (>>> with postive A, same rules as number 3 above. 6. For unsigned right shift operator (>>> with negative A, do it conventional way: a. find the 2’s compliment of negative A b. right shift by B bits, inserting 0 on the left c. find the decimal of above.

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