I have started giving mock tests available on the internet. One problem that I see is that I get stuck in the questions involving shift operators. While learning about shift operators I used calculators to get to the right answer. But calculators are not allowed in the final exam. I get nervous when I come across big numbers carrying only 0's and 1's. Is there any short cut method of solving questions involving shift operators? For example: -4 >>> 3. Could someone tell me all the steps in solving this problem?
Take it easy. If you look in the Java™ Tutorial, you find this:
Certain operators tend to appear more frequently than others; for example, the assignment operator "=" is far more common than the unsigned right shift operator ">>>".
So you needn't expect to find lots of questions about shift operators.
You usually use them on ints mainly because ints are the most popular type of number, but I shall show you how they work on bytes, mainly because it is quicker to write 8 1s than 32!
1: The left-shift operator. << n
This moves the whole thing n bits to the left; any bits which pass the most significant bits disappear into cyber-limbo and any gaps on the right are filled with 0s. So 21 << 2 is
21 = 00010101
Two to the left = 010101--
Fill in the 0s = 01010100
Convert to decimal = 84.
84 / 2 = 4, and 4 = 2^2. There's your 2 back.
Left-shift 1 bit is equivalent to multiplying by 2, two bits is multiplying by 4, and three bits is multiplying by 8. The normal rules about overflow errors apply; 21 << 3 in 32 bits is 168 but in 8 bits it's -88. -33 << 2 in 32 bits is -132, in 8 bits it's +124.
2: The right-shift operator >> n This moves the whole thing n bits to the right; any bits which pass the least significant bit fall off the edge of the world and it fills in from the left with the same bit that the original most significant bit was. 21 >> 2 is
21 = 00010101
2 to the right gives --000101
Fill in with 0 gives 00000101
and that's 5
21 / 5 in integer arithmetic is 4, 4 = 2^2, and there's your 2 back.
Not 168, but -88 is 10101000.
2 to the right gives --101010
Fill in with 1s this time gives 11101010
11101010 is -22.
-88 / -22 is 4, and 4 is 2^2.
So right-shift by 1 is the same as dividing by 2, two bits to the right is the same as dividing by 4, three bits is dividing by 8, etc. Note that the sign is always maintained and there is no risk of overflow, but if you shift too many places you will always get -1 from negative numbers and 0 from positive numbers.
Both the right-shift and left-shift operators are usually described as being "signed."
3: The unsigned right shift >>> n This is the most complicated of the three. It moves the bits n places to the right, and any bits which pass the least significant bit float off into the wild blue yonder, but it fills in from the left with 0s. So it always returns a positive answer, whether the original answer was positive or negative. For positive numbers it has exactly the same effect as the ordinary signed right-shift, so I won't draw it For a negative number, eg -88 >>> 2, this is what happens
-88 is 10101000
Two to the right gives --101010
Fill in with ZEROES from the left: 00101010
00101010 is 42.
Remember that is you multiply 42 by 4 in 8 bits you get -88. I am afraid I can't think of a simple way to calculate the values you get from a >>> operator.
I hope this is of some help to you And it was a nice interesting question to answer, too.
Joined: Oct 13, 2005
Try this little programIt works; I have just tried it.
But for the last it converts to an int. See what happens; it returns what you get from the right-shift as an int.
I am sorry I took long to reply. I appreciate from the bottom of my heart the effort you all have put in to help me understand shifting. I tried some questions manually and was able to solve. There is still slight problem with >>> for negative numbers such as -16>>>3, the 32 bits are over-whelming. For instance...
Simple, use A.P series most of the calculation can be done in my mind if one is good with mathematics in very few seconds.
say your bits look like , 11111111 1111111 11111100
As per your convinience take From either right to left or vice versa. Say from right,
Count the no. of 1's until zero is encountered.Therefore consecutive 1's is 22. So, the above case A.P series can be written as
2^23 + 2^22 +.....2^2+ 0 + 0 --------------------- 22's 1 (but mind it all are 1's are consecutive)
just apply the formula for sum= n/2(2*a+(n-1)/d)
PS: you may have to work little bit more if alternately bits are there but this method is still very fast. [ March 23, 2008: Message edited by: Shivit Agarwal ]
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