The operator which is used to shift the bit patterns right or left is called shift operator. The general form of shift expression is as follows:
left_operand op n
where,
left_operand ➜ left operand that can be of integral type.
op ➜ left shift or right shift operator.
n ➜ number of bit positions to be shifted. It must be of type int only.
For example:
10 << 2
Here, the value 10 is operand, << is left shift operator, 2 is the number of bit positions to be shifted.
There are two types of shift operators.
1. Left shift operator.
2. Right shift operator.
The operator that shifts the bits of number towards left by n number of bit positions is called left shift operator in Java. This operator is represented by a symbol <<, read as double less than.
if we write x << n, it means that the bits of x will be shifted towards left by n positions.
Let us take an example to understand the concept of the left shift operator. If int x = 20. Calculate x value if x << 3.
The value of x is 20 = 0 0 0 1 0 1 0 0 (binary format). Now x << 3 will shift the bits of x towards left by 3 positions. Due to which leftmost 3 bits will be lost. Hence, after shifting, bits of x is 1 0 1 0 0 0 0 0 that is 160 in decimal form. The bit pattern after the left shift is shown in the below figure:
Key point:
Shifting a value to the left, n bits, is equivalent to multiplying that value by 2^n.
For example:
In the above expression, n = 3. So, 20 * 2^3 = 20 * 8 = 160.
Let's take another example. If a = 45, calculate a value if a << 1.
The value 45 is represented in binary format as 0 0 1 0 1 1 0 1. If this value is shifted towards left by one position, we will get 0 1 0 1 1 0 1 0. This number in decimal form is nothing but 90 that is twice of 45. i.e. 45 * 2^1 = 90.
The operator that shifts the bits of number towards the right by n number of bit positions is called right shift operator in Java. The right shift operator in java is represented by a symbol >>, read as double greater than.
If we write x >> n, it means that the bits of x will be shifted towards right by n positions. There are two types of right shift operators in java:
1. Signed right shift operator (>>)
2. Unsigned right shift operator (>>>).
Both of these operators shifts bits of number towards right by n number of bit positions.
If the number is positive, the leftmost position is filled with 0. If the number is negative, the leftmost position is filled with 1. The signed shift operator uses the same sign as used in the number before shifting of bits.
Let's understand the concept of right shift operator with the help of an example. If int x = 10 then calculate x >> 3 value.
The value of x is 10 = 0 0 0 0 1 0 1 0. Since the number is positive, the leftmost bit position will be filled with 0. Now x >> 3 will shift the bits of x towards the right by 3 positions. The rightmost 3 bits will be lost due to shifting. Hence, after shifting, bits of x is 0 0 0 0 0 0 0 1 that is 1 in decimal form. The bit pattern after the right shift is shown in the below figure:
You will notice in this example that after performing right shift operation on a positive number 10, we get a positive value 1 as a result. Similarly, if we perform right shift operation on a negative number, again we will get a negative value only.
Key point:
Shifting a value to the right is equivalent to dividing the number by 2^n.
For example:
1. int i = 10;
int result = i >> 3; // result = 1
In this expression, 10 / 2^3 = 10 / 8 = 1.
2. int i =20;
int result = i >> 2; // result = 5.
In this expression, 20 / 2^2 = 5.
3. int i = 20
int result = i >> 2; // result = 5.
In this expression, 20 / 2^2 = 5.
Let's take an example that is based on right shifting on a negative number. If int x = 10 then calculate x >> 2 value.
The value of x is 10 that is a negative number. So, we will use 2's complement system to represent the negative numbers. Follow all the steps given below in the figure.
1. +10 in 8bit form is 0 0 0 0 1 0 1 0.
2. Obtain the 1's complement of 0 0 0 0 1 0 1 0. The 1's complement can be obtained by simply complementing each bit of the number, that is, by changing all 0s to 1s and all 1s to 0s. After taking complement of 0 0 0 0 1 0 1 0, the result in 1's complement form is 1 1 1 1 0 1 0 1.
3. Now add 1 to get 2's complement form. After adding 1 in LSB (Least Significant Bit), the result is 1 1 1 1 0 1 1 0. This is in 2's complement form. Here, 1 in MSB (Most Significant Bit) represents negative number. Thus, the 2's complement form of 10 is 1 1 1 1 0 1 1 0.
4. Since the number is negative, the leftmost position is filled with two 1s. Now x >> 2 will shift the bits of x towards the right by 2 positions. The rightmost 2 bits will be lost due to shifting. Hence, after shifting, bits of x in 2's complement form is 1 1 1 1 1 1 0 1.
5. Now convert 2's complement bits into 1's complement bits. Subtract 1 in LSB to convert 2's complement form into 1's complement. The result in 1's complement form is 1 1 1 1 1 1 0 0.
6. Take the complement of each bit to get original form of number after shifting. So, the result in original form is 0 0 0 0 0 0 1 1. This is nothing but 3 in decimal form.
Let's verify the above result by creating a program in java.
Program source code 1:
left_operand op n
where,
left_operand ➜ left operand that can be of integral type.
op ➜ left shift or right shift operator.
n ➜ number of bit positions to be shifted. It must be of type int only.
For example:
10 << 2
Here, the value 10 is operand, << is left shift operator, 2 is the number of bit positions to be shifted.
There are two types of shift operators.
1. Left shift operator.
2. Right shift operator.
Left Shift Operator in Java
The operator that shifts the bits of number towards left by n number of bit positions is called left shift operator in Java. This operator is represented by a symbol <<, read as double less than.
if we write x << n, it means that the bits of x will be shifted towards left by n positions.
Let us take an example to understand the concept of the left shift operator. If int x = 20. Calculate x value if x << 3.
The value of x is 20 = 0 0 0 1 0 1 0 0 (binary format). Now x << 3 will shift the bits of x towards left by 3 positions. Due to which leftmost 3 bits will be lost. Hence, after shifting, bits of x is 1 0 1 0 0 0 0 0 that is 160 in decimal form. The bit pattern after the left shift is shown in the below figure:
Shifting a value to the left, n bits, is equivalent to multiplying that value by 2^n.
For example:
In the above expression, n = 3. So, 20 * 2^3 = 20 * 8 = 160.
Let's take another example. If a = 45, calculate a value if a << 1.
The value 45 is represented in binary format as 0 0 1 0 1 1 0 1. If this value is shifted towards left by one position, we will get 0 1 0 1 1 0 1 0. This number in decimal form is nothing but 90 that is twice of 45. i.e. 45 * 2^1 = 90.
Right Shift Operator in Java
The operator that shifts the bits of number towards the right by n number of bit positions is called right shift operator in Java. The right shift operator in java is represented by a symbol >>, read as double greater than.
If we write x >> n, it means that the bits of x will be shifted towards right by n positions. There are two types of right shift operators in java:
1. Signed right shift operator (>>)
2. Unsigned right shift operator (>>>).
Both of these operators shifts bits of number towards right by n number of bit positions.
Signed Right Shift Operator in Java
The signed right shift operator >> shifts bits of the number towards the right and also reserve sign bit, which is leftmost bit. A sign bit represents the sign of a number. If the sign bit is 0 then it represents a positive number. If sign bit is 1, it represents a negative number.
If the number is positive, the leftmost position is filled with 0. If the number is negative, the leftmost position is filled with 1. The signed shift operator uses the same sign as used in the number before shifting of bits.
Let's understand the concept of right shift operator with the help of an example. If int x = 10 then calculate x >> 3 value.
The value of x is 10 = 0 0 0 0 1 0 1 0. Since the number is positive, the leftmost bit position will be filled with 0. Now x >> 3 will shift the bits of x towards the right by 3 positions. The rightmost 3 bits will be lost due to shifting. Hence, after shifting, bits of x is 0 0 0 0 0 0 0 1 that is 1 in decimal form. The bit pattern after the right shift is shown in the below figure:
You will notice in this example that after performing right shift operation on a positive number 10, we get a positive value 1 as a result. Similarly, if we perform right shift operation on a negative number, again we will get a negative value only.
Key point:
Shifting a value to the right is equivalent to dividing the number by 2^n.
For example:
1. int i = 10;
int result = i >> 3; // result = 1
In this expression, 10 / 2^3 = 10 / 8 = 1.
2. int i =20;
int result = i >> 2; // result = 5.
In this expression, 20 / 2^2 = 5.
3. int i = 20
int result = i >> 2; // result = 5.
In this expression, 20 / 2^2 = 5.
Let's take an example that is based on right shifting on a negative number. If int x = 10 then calculate x >> 2 value.
The value of x is 10 that is a negative number. So, we will use 2's complement system to represent the negative numbers. Follow all the steps given below in the figure.
1. +10 in 8bit form is 0 0 0 0 1 0 1 0.
2. Obtain the 1's complement of 0 0 0 0 1 0 1 0. The 1's complement can be obtained by simply complementing each bit of the number, that is, by changing all 0s to 1s and all 1s to 0s. After taking complement of 0 0 0 0 1 0 1 0, the result in 1's complement form is 1 1 1 1 0 1 0 1.
3. Now add 1 to get 2's complement form. After adding 1 in LSB (Least Significant Bit), the result is 1 1 1 1 0 1 1 0. This is in 2's complement form. Here, 1 in MSB (Most Significant Bit) represents negative number. Thus, the 2's complement form of 10 is 1 1 1 1 0 1 1 0.
4. Since the number is negative, the leftmost position is filled with two 1s. Now x >> 2 will shift the bits of x towards the right by 2 positions. The rightmost 2 bits will be lost due to shifting. Hence, after shifting, bits of x in 2's complement form is 1 1 1 1 1 1 0 1.
5. Now convert 2's complement bits into 1's complement bits. Subtract 1 in LSB to convert 2's complement form into 1's complement. The result in 1's complement form is 1 1 1 1 1 1 0 0.
6. Take the complement of each bit to get original form of number after shifting. So, the result in original form is 0 0 0 0 0 0 1 1. This is nothing but 3 in decimal form.
Let's verify the above result by creating a program in java.
Program source code 1:

package shiftOperator;
public class SignedRightShiftExample {
public static void main(String[] args) {
int x = 10, c = 0;
c = x >> 2;
System.out.println("x >> 2 = " +c);
}
}
 Output:
x >> 2 = 3
Unsigned Right Shift Operator in Java
The unsigned right shift operator in java performs nearly the same operation as the signed right shift operator in java. The unsigned right shift operator is represented by a symbol >>>, read as triple greater than.
The unsigned right shift operator always fills the leftmost position with 0s because the value is not signed. Since it always stores 0 in the sign bit, it is also called zero fill right shift operator in java.
If you will apply the unsigned right shift operator >>> on a positive number, it will give the same result as that of >>. But in the case of negative numbers, the result will be positive because the signed bit is replaced by 0.
Key points:
Both signed right shift operator >> and unsigned right shift operator >>> are used to shift bits towards right. The only difference between them is that >> preserve sign bits whereas >>> does not preserve sign bit. It always fills 0 in the sign bit whether number is positive or negative.
Let us create a program to observe the effect of various shift operators.
Program source code 2:

package shiftOperator;
public class UnsignedRightShiftExample {
public static void main(String[] args) {
int x = 10;
int y = 10;
System.out.println("x >> 1 = " + (x >> 1));
System.out.println("x >>> 1 = " + (x >>> 1));
System.out.println("y >> 2 = " + (y >> 2));
System.out.println("y >>> 2 = " + (y >>> 2));
}
}

Output:
x >> 1 = 5
x >>> 1 = 5
y >> 2 = 3
y >>> 2 = 1073741821