Complete Reference for Bitwise Operators in Programming/Coding

There exists no programming language that doesn't use Bit Manipulations. Bit manipulation is all about these bitwise operations. They improve the efficiency of programs by being primitive, fast actions. There are different bitwise operations used in bit manipulation. These Bitwise Operators operate on the individual bits of the bit patterns. Bit operations are fast and can be used in optimizing time complexity.

Some common bit operators are:

1. Bitwise AND Operator (&)

The bitwise AND operator is denoted using a single ampersand symbol, i.e. &. The & operator takes two equal-length bit patterns as parameters. The two-bit integers are compared. If the bits in the compared positions of the bit patterns are 1, then the resulting bit is 1. If not, it is 0.

Example: 

Take two bit values X and Y, where X = 7= (111)₂ and Y = 4 = (100)₂ . Take Bitwise and of both X & y

Implementation of AND operator:

#include <bits/stdc++.h>
using namespace std;

int main()
{

    int a = 7, b = 4;
    int result = a & b;
    cout << result << endl;

    return 0;
}

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main (String[] args) {
        int a = 7, b = 4;
          int result = a & b;
          System.out.println(result);
    }
}

// This code is contributed by lokeshmvs21.

a = 7
b = 4
result = a & b
print(result)
# This code is contributed by akashish__

using System;

public class GFG{

    static public void Main (){
      int a = 7, b = 4;
      int result = a & b;
      Console.WriteLine(result);
    }
}

// This code is contributed by akashish__

let a = 7, b = 4;
let result = a & b;
console.log(result);
// This code is contributed by akashish__

4

Time Complexity: O(1) 
Auxiliary Space: O(1)

2. ​Bitwise OR Operator (|)

The | Operator takes two equivalent length bit designs as boundaries; if the two bits in the looked-at position are 0, the next bit is zero. If not, it is 1.

Example: 

Take two bit values X and Y, where X = 7= (111)₂ and Y = 4 = (100)₂ . Take Bitwise OR of both X, y

Explanation: On the basis of truth table of bitwise OR operator we can conclude that the result of 

1 | 1  = 1
1 | 0 = 1
0 | 1 = 1
0 | 0 = 0

We used the similar concept of bitwise operator that are show in the image.

Implementation of OR operator:

#include <bits/stdc++.h>
using namespace std;

int main()
{

    int a = 12, b = 25;
    int result = a | b;
    cout << result;

    return 0;
}

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int a = 12, b = 25;
        int result = a | b;
        System.out.println(result);
    }
}

a = 12
b = 25
result = a | b
print(result)

# This code is contributed by garg28harsh.

using System;

public class GFG{

    static public void Main (){
        int a = 12, b = 25;
        int result = a | b;
        Console.WriteLine(result);
    }
}
// This code is contributed by akashish__

    let a = 12, b = 25;
    let result = a | b;
    document.write(result);
      
// This code is contributed by garg28harsh.

29

Time Complexity: O(1) 
Auxiliary Space: O(1)

3. ​Bitwise XOR Operator (^)

The ^ operator (also known as the XOR operator) stands for Exclusive Or. Here, if bits in the compared position do not match their resulting bit is 1. i.e, The result of the bitwise XOR operator is 1 if the corresponding bits of two operands are opposite, otherwise 0.

Example: 

Take two bit values X and Y, where X = 7= (111)₂ and Y = 4 = (100)₂ . Take Bitwise and of both X & y

Explanation: On the basis of truth table of bitwise XOR operator we can conclude that the result of 

1 ^ 1  = 0
1 ^ 0 = 1
0 ^ 1 = 1
0 ^ 0 = 0

We used the similar concept of bitwise operator that are show in the image.

Implementation of XOR operator:

#include <iostream>
using namespace std;

int main()
{

    int a = 12, b = 25;
    cout << (a ^ b);
    return 0;
}

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int a = 12, b = 25;
        int result = a ^ b;
        System.out.println(result);
    }
}

// This code is contributed by garg28harsh.

a = 12
b = 25
result = a ^ b
print(result)

# This code is contributed by garg28harsh.

// C# Code

using System;

public class GFG {

    static public void Main()
    {

        // Code
        int a = 12, b = 25;
        int result = a ^ b;
        Console.WriteLine(result);
    }
}

// This code is contributed by lokesh

let a = 12;
let b = 25;
console.log((a ^ b));
// This code is contributed by akashish__

21

Time Complexity: O(1) 
Auxiliary Space: O(1)

4. ​Bitwise NOT Operator (!~)

All the above three bitwise operators are binary operators (i.e, requiring two operands in order to operate). Unlike other bitwise operators, this one requires only one operand to operate.

The bitwise Not Operator takes a single value and returns its one’s complement. The one’s complement of a binary number is obtained by toggling all bits in it, i.e, transforming the 0 bit to 1 and the 1 bit to 0.

Example: 

Take two bit values X and Y, where X = 5= (101)₂ . Take Bitwise NOT of X.

Explanation: On the basis of truth table of bitwise NOT operator we can conclude that the result of 

~1  = 0
~0 = 1

We used the similar concept of bitwise operator that are show in the image.

Implementation of NOT operator:

#include <iostream>
using namespace std;

int main()
{

    int a = 0;
    cout << "Value of a without using NOT operator: " << a;
    cout << "\nInverting using NOT operator (with sign bit): " << (~a);
    cout << "\nInverting using NOT operator (without sign bit): " << (!a);

    return 0;
}

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
  public static void main(String[] args)
  {
    int a = 0;
    System.out.println(
      "Value of a without using NOT operator: " + a);
    System.out.println(
      "Inverting using NOT operator (with sign bit): "
      + (~a));
    if (a != 1)
      System.out.println(
      "Inverting using NOT operator (without sign bit): 1");
    else
      System.out.println(
      "Inverting using NOT operator (without sign bit): 0");
  }
}

// This code is contributed by lokesh.

a = 0
print("Value of a without using NOT operator: " , a)
print("Inverting using NOT operator (with sign bit): " , (~a))
print("Inverting using NOT operator (without sign bit): " , int(not(a)))
#  This code is contributed by akashish__

using System;

public class GFG {

  static public void Main()
  {

    int a = 0;
    Console.WriteLine(
      "Value of a without using NOT operator: " + a);
    Console.WriteLine(
      "Inverting using NOT operator (with sign bit): "
      + (~a));
    if (a != 1)
      Console.WriteLine(
      "Inverting using NOT operator (without sign bit): 1");
    else
      Console.WriteLine(
      "Inverting using NOT operator (without sign bit): 0");
  }
}

// This code is contributed by akashish__

    let a =0;
    document.write("Value of a without using NOT operator: " + a);
    document.write( "Inverting using NOT operator (with sign bit): " + (~a));
    if(!a)
    document.write( "Inverting using NOT operator (without sign bit): 1" );
    else
    document.write( "Inverting using NOT operator (without sign bit): 0" );

    

Value of a without using NOT operator: 0
Inverting using NOT operator (with sign bit): -1
Inverting using NOT operator (without sign bit): 1

Time Complexity: O(1) 
Auxiliary Space: O(1)

5. Left-Shift (<<)

The left shift operator is denoted by the double left arrow key (<<). The general syntax for left shift is shift-expression << k. The left-shift operator causes the bits in shift expression to be shifted to the left by the number of positions specified by k. The bit positions that the shift operation has vacated are zero-filled. 

Note: Every time we shift a number towards the left by 1 bit it multiply that number by 2.

Example:

Input: Left shift of 5 by 1.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 1)
 

Output: 10
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 01010_2, Which is equivalent to 10

Input: Left shift of 5 by 2.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 2)

Output: 20
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 10100_2, Which is equivalent to 20

Input: Left shift of 5 by 3.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 3)

Output: 40
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 01000_2, Which is equivalent to 40

Implementation of Left shift operator:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    unsigned int num1 = 1024;

    bitset<32> bt1(num1);
    cout << bt1 << endl;

    unsigned int num2 = num1 << 1;
    bitset<32> bt2(num2);
    cout << bt2 << endl;

    unsigned int num3 = num1 << 2;
    bitset<16> bitset13{ num3 };
    cout << bitset13 << endl;
}

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
  public static void main(String[] args)
  {
    int num1 = 1024;

    String bt1 = Integer.toBinaryString(num1);
    bt1 = String.format("%32s", bt1).replace(' ', '0');
    System.out.println(bt1);

    int num2 = num1 << 1;
    String bt2 = Integer.toBinaryString(num2);
    bt2 = String.format("%32s", bt2).replace(' ', '0');
    System.out.println(bt2);

    int num3 = num1 << 2;
    String bitset13 = Integer.toBinaryString(num3);
    bitset13 = String.format("%16s", bitset13)
      .replace(' ', '0');
    System.out.println(bitset13);
  }
}

// This code is contributed by akashish__

# Python code for the above approach

num1 = 1024

bt1 = bin(num1)[2:].zfill(32)
print(bt1)

num2 = num1 << 1
bt2 = bin(num2)[2:].zfill(32)
print(bt2)

num3 = num1 << 2
bitset13 = bin(num3)[2:].zfill(16)
print(bitset13)

# This code is contributed by Prince Kumar

using System;

class GFG {
  public static void Main(string[] args)
  {
    int num1 = 1024;

    string bt1 = Convert.ToString(num1, 2);
    bt1 = bt1.PadLeft(32, '0');
    Console.WriteLine(bt1);

    int num2 = num1 << 1;
    string bt2 = Convert.ToString(num2, 2);
    bt2 = bt2.PadLeft(32, '0');
    Console.WriteLine(bt2);

    int num3 = num1 << 2;
    string bitset13 = Convert.ToString(num3, 2);
    bitset13 = bitset13.PadLeft(16, '0');
    Console.WriteLine(bitset13);
  }
}

// This code is contributed by akashish__

// JavaScript code for the above approach

let num1 = 1024;

let bt1 = num1.toString(2).padStart(32, '0');
console.log(bt1);

let num2 = num1 << 1;
let bt2 = num2.toString(2).padStart(32, '0');
console.log(bt2);

let num3 = num1 << 2;
let bitset13 = num3.toString(2).padStart(16, '0');
console.log(bitset13);

00000000000000000000010000000000
00000000000000000000100000000000
0001000000000000

Time Complexity: O(1) 
Auxiliary Space: O(1)

6. Right-Shift (>>)

The right shift operator is denoted by the double right arrow key (>>). The general syntax for the right shift is "shift-expression >> k". The right-shift operator causes the bits in shift expression to be shifted to the right by the number of positions specified by k. For unsigned numbers, the bit positions that the shift operation has vacated are zero-filled. For signed numbers, the sign bit is used to fill the vacated bit positions. In other words, if the number is positive, 0 is used, and if the number is negative, 1 is used.

Note: Every time we shift a number towards the right by 1 bit it divides that number by 2.

Example:

Input: Left shift of 5 by 1.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 1)

Output: 10
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 01010_2, Which is equivalent to 10

Input: Left shift of 5 by 2.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 2)

Output: 20
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 10100_2, Which is equivalent to 20

Input: Left shift of 5 by 3.
Binary representation of 5 = 00101 and Left shift of 00101₂ by 1 (i.e, 00101 << 3)

Output: 40
Explanation: All bit of 5 will be shifted by 1 to left side and this result in 01000_2, Which is equivalent to 40

Implementation of Right shift operator:

#include <bitset>
#include <iostream>

using namespace std;

int main()
{
    unsigned int num1 = 1024;

    bitset<32> bt1(num1);
    cout << bt1 << endl;

    unsigned int num2 = num1 >> 1;
    bitset<32> bt2(num2);
    cout << bt2 << endl;

    unsigned int num3 = num1 >> 2;
    bitset<16> bitset13{ num3 };
    cout << bitset13 << endl;
}

// Java code for the above approach

class GFG {
    public static void main(String[] args)
    {
        int num1 = 1024;

        String bt1
            = String
                  .format("%32s",
                          Integer.toBinaryString(num1))
                  .replace(' ', '0');
        System.out.println(bt1);

        int num2 = num1 >> 1;
        String bt2
            = String
                  .format("%32s",
                          Integer.toBinaryString(num2))
                  .replace(' ', '0');
        System.out.println(bt2);

        int num3 = num1 >> 2;
        String bitset13
            = String
                  .format("%16s",
                          Integer.toBinaryString(num3))
                  .replace(' ', '0');
        System.out.println(bitset13);
    }
}

// This code is contributed by ragul21

num1 = 1024
bt1 = bin(num1)[2:].zfill(32)
print(bt1)

num2 = num1 >> 1
bt2 = bin(num2)[2:].zfill(32)
print(bt2)

num3 = num1 >> 2
bitset13 = bin(num3)[2:].zfill(16)
print(bitset13)

using System;

class Program
{
    static void Main()
    {
        int num1 = 1024;
        
        // Right shift by 1
        int num2 = num1 >> 1;
        
        // Right shift by 2
        int num3 = num1 >> 2;

        // Print binary representations
        string bt1 = Convert.ToString(num1, 2).PadLeft(32, '0');
        string bt2 = Convert.ToString(num2, 2).PadLeft(32, '0');
        string bitset13 = Convert.ToString(num3, 2).PadLeft(16, '0');

        Console.WriteLine(bt1);
        Console.WriteLine(bt2);
        Console.WriteLine(bitset13);
    }
}

// JavaScript code for the above approach

let num1 = 1024;

let bt1 = num1.toString(2).padStart(32, '0');
console.log(bt1);

let num2 = num1 >> 1;
let bt2 = num2.toString(2).padStart(32, '0');
console.log(bt2);

let num3 = num1 >> 2;
let bitset13 = num3.toString(2).padStart(16, '0');
console.log(bitset13);
// akashish__

00000000000000000000010000000000
00000000000000000000001000000000
0000000100000000

Time Complexity: O(1) 
Auxiliary Space: O(1)

Application of BIT Operators

• Bit operations are used for the optimization of embedded systems.
• The Exclusive-or operator can be used to confirm the integrity of a file, making sure it has not been corrupted, especially after it has been in transit.
• Bitwise operations are used in Data encryption and compression.
• Bits are used in the area of networking, framing the packets of numerous bits which are sent to another system generally through any type of serial interface.
• Digital Image Processors use bitwise operations to enhance image pixels and to extract different sections of a microscopic image.

kartik

Follow

Improve

Article Tags :

• Bit Magic
• DSA
• Programming
• Bitwise-XOR
• Bitwise-OR
• Bitwise-AND
• Bitwise Operators

+3 More

Practice Tags :

• Bit Magic