What are bitwise shift (bit-shift) operators and how do they work?

Question

I\'ve been attempting to learn C in my spare time, and other languages (C#, Java, etc.) have the same concept (and often the same operators) ...

What I\'m wondering

Answer 1

The Bitwise operators are used to perform operations a bit-level or to manipulate bits in different ways. The bitwise operations are found to be much faster and are some times used to improve the efficiency of a program. Basically, Bitwise operators can be applied to the integer types: long, int, short, char and byte.

Bitwise Shift Operators

They are classified into two categories left shift and the right shift.

• Left Shift(<<): The left shift operator, shifts all of the bits in value to the left a specified number of times. Syntax: value << num. Here num specifies the number of position to left-shift the value in value. That is, the << moves all of the bits in the specified value to the left by the number of bit positions specified by num. For each shift left, the high-order bit is shifted out (and ignored/lost), and a zero is brought in on the right. This means that when a left shift is applied to 32-bit compiler, bits are lost once they are shifted past bit position 31. If the compiler is of 64-bit then bits are lost after bit position 63.

Output: 6, Here the binary representation of 3 is 0...0011(considering 32-bit system) so when it shifted one time the leading zero is ignored/lost and all the rest 31 bits shifted to left. And zero is added at the end. So it became 0...0110, the decimal representation of this number is 6.

• In the case of a negative number:

Output: -2, In java negative number, is represented by 2's complement. SO, -1 represent by 2^32-1 which is equivalent to 1....11(Considering 32-bit system). When shifted one time the leading bit is ignored/lost and the rest 31 bits shifted to left and zero is added at the last. So it becomes, 11...10 and its decimal equivalent is -2. So, I think you get enough knowledge about the left shift and how its work.

• Right Shift(>>): The right shift operator, shifts all of the bits in value to the right a specified of times. Syntax: value >> num, num specifies the number of positions to right-shift the value in value. That is, the >> moves/shift all of the bits in the specified value of the right the number of bit positions specified by num. The following code fragment shifts the value 35 to the right by two positions:

Output: 8, As a binary representation of 35 in a 32-bit system is 00...00100011, so when we right shift it two times the first 30 leading bits are moved/shifts to the right side and the two low-order bits are lost/ignored and two zeros are added at the leading bits. So, it becomes 00....00001000, the decimal equivalent of this binary representation is 8. Or there is a simple mathematical trick to find out the output of this following code: To generalize this we can say that, x >> y = floor(x/pow(2,y)). Consider the above example, x=35 and y=2 so, 35/2^2 = 8.75 and if we take the floor value then the answer is 8.

Output:

But remember one thing this trick is fine for small values of y if you take the large values of y it gives you incorrect output.

• In the case of a negative number: Because of the negative numbers the Right shift operator works in two modes signed and unsigned. In signed right shift operator (>>), In case of a positive number, it fills the leading bits with 0. And In case of a negative number, it fills leading bits with 1. To keep the sign. This is called 'sign extension'.

Output: -5, As I explained above the compiler stores the negative value as 2's complement. So, -10 is represented as 2^32-10 and in binary representation considering 32-bit system 11....0110. When we shift/ move one time the first 31 leading bits got shifted in the right side and the low-order bit got lost/ignored. So, it becomes 11...0011 and the decimal representation of this number is -5 (How I know the sign of number? because the leading bit is 1). It is interesting to note that if you shift -1 right, the result always remains -1 since sign extension keeps bringing in more ones in the high-order bits.

• Unsigned Right Shift(>>>): This operator also shifts bits to the right. The difference between signed and unsigned is the latter fills the leading bits with 1 if the number is negative and the former fills zero in either case. Now the question arises why we need unsigned right operation if we get the desired output by signed right shift operator. Understand this with an example, If you are shifting something that does not represent a numeric value, you may not want sign extension to take place. This situation is common when you are working with pixel-based values and graphics. In these cases, you will generally want to shift a zero into the high-order bit no matter what it's the initial value was.

Output: 2147483647, Because -2 is represented as 11...10 in a 32-bit system. When we shift the bit by one, the first 31 leading bit is moved/shifts in right and the low-order bit is lost/ignored and the zero is added to the leading bit. So, it becomes 011...1111 (2^31-1) and its decimal equivalent is 2147483647.