C plus plus bitwise operators

Table of Contents

Most high-level language does not allow a bitwise operation because it is a machine level task. The bit is manipulated in memory to get the desired results using bitwise operators.

Bitwise AND

A bitwise operation happens between two bits. A number is converted into a bit string and then bitwise operations performed on them.

The $bitwise \hspace{3px} AND$ works similarly as $logical \hspace{3px} AND$ . All output combination is show for $bitwise \hspace{3px} AND$ below.

Let $A$ and $B$ be two variables with some one bit value, then:

A	B	Output
1	1	1
1	0	0
0	1	0
0	0	0

The only time output is $1$ when both $A$ and $B$ are $1$ .

For example, $bitwise \hspace{3px} AND$ with $12$ and $7$ will result in following:

\begin{aligned}&1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}0 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}12\\ \\ &0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}7\\ \\ &--------------\\ &0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}0 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}4\\ &--------------\end{aligned} ﻿

The result is number $4$ .

Example Program: Bitwise AND

We now illustrate the correctness of the $bitwise \hspace{3px}AND$ through a program.

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{
    
    //Variable Declaration
    
    unsigned int a, b;
    unsigned int result;
    //Variable initialization
    
    a = 12;
    b = 7;
       
    result = a & b;
    //printing the results
       
    cout << " Output = " << " " << result << endl;
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = 4

Bitwise OR

The $bitwise \hspace{3px} OR$ is similar to $logical \hspace{3px} OR$ you learned earlier, except that it works at bit level. When a $bitwise \hspace{3px}OR$ is performed on two bits – $A$ and $B$ , then the result is $1$ if the bit value of $A$ or $B$ is $1$ . The bitwise operations are possible for more than two bits.

All combination of bit values is given below for $bitwise \hspace{3px} OR$ .

A	B	Output
1	1	1
1	0	1
0	1	1
0	0	0

For example,

If you want to perform $bitwise \hspace{3px} OR$ operation on $A = 14$ and $B = 3$ then

\begin{aligned}&1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}14\\ \\ &0 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}3\\ \\ &--------------\\ &1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}15\\ &--------------\end{aligned} ﻿

Therefore, the output is $15$ .

Example Program: Bitwise OR

To verify the output from above example, we will write a program that computes using $bitwise \hspace{3px} OR$ .

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{
    //Variable declarations
    
    int a,b,result;
    
    //Variable initialization
    
    a = 14;
    b = 3;
    result = 0;
    
    // Bitwise OR operation
    
    result = a | b;
    
    //printing the results
    
    cout << "Output =" << " " << result << endl;
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = 15

The output is what we expected.

Bitwise XOR (Exclusive-OR)

The $bitwise \hspace{3px} XOR$ is a special kind of logical operation where two single bit inputs – $A$ and $B$ gives an output of $1$ if either of them if is $1$ . Otherwise, the result is $0$ .

All combination of two inputs and a single output is given for $bitwise \hspace{3px} XOR$ below.

A	B	Output
1	1	0
1	0	1
0	1	1
0	0	0

Note how the output is $1$ when either input is $1$ .

Let us see a manual example for $bitwise \hspace{3px} XOR$ and later create a program that performs $bitwise \hspace{3px} XOR$ to verify our output.

If $A = 24$ and $B = 45$ then $bitwise \hspace{3px} XOR$ will result in following result.

\begin{aligned}&0 \hspace{5px}0 \hspace{5px}0\hspace{5px}1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}0\hspace{5px}0 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}24\\ \\ &0 \hspace{5px}0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1\hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}45\\ \\ &-------------------\\ &0 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}53\\ &-------------------\end{aligned} ﻿

Therefore, the output $0011 \hspace{2px} 0101$ is binary equivalent of $53$ .

Example Program: Bitwise XOR

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{
    //Variable declarations
    
    int a,b,result;
    
    //Variable initialization
    
    a = 24;
    b = 45;
    result = 0;
    
    // Bitwise OR operation
    
    result = a ^ b;
    
    //printing the results
    
    cout << "Output =" << " " << result << endl;
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = 53

Left Shift Operator (<<)

The value of the variable can be modified by changing the bits. This is achieved using bitwise shift operators.

The $left \hspace{3px}shift\hspace{3px} operator$ modifies the value of variable by shifting the leftmost bits.

For example,

If $A = 23$ and you want to change the value of $A$ using $left \hspace{3px}shift\hspace{3px} operator$ ,

\begin{aligned}&0 \hspace{5px}0 \hspace{5px}0\hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}23 \\ \\ &0 \hspace{5px}0 \hspace{5px}1\hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}46 \hspace{5px}after\hspace{5px}1 \hspace{5px}bit \hspace{5px}left\hspace{5px} shift \end{aligned}

Therefore, after shifting a single bit to the left, we get $46$ .

Example Program: Left Shift

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{ 
    //Variable Declaration
     int A;
     
     // Variable initialization
     
     A = 23;
     
     //Bitwise Left Shift 
     
     A = A << 1;
     
     // Printing the results
     
     cout << "Output = " << " " << A << endl;
     
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = 46

It is not necessary that you have to shift only one bit, you can shift more than one to archive the desired results.

Right Shift Operator(>>)

The $right\hspace{3px} shift \hspace{3px}operator$ is similar to left shift operator and it modify the value of a variable by shifting the rightmost bits.

For example,

If $A = 23$ and you want to shift rightmost bit then,

\begin{aligned}&0 \hspace{5px}0 \hspace{5px}0\hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}23 \\ \\ &0 \hspace{5px}0 \hspace{5px}0\hspace{5px}0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\backslash \backslash Binary \hspace{5px}value\hspace{5px} of \hspace{5px}11\hspace{5px}after\hspace{5px}1 \hspace{5px}bit \hspace{5px}right\hspace{5px} shift \end{aligned}

After shift $1$ bit to the right we get $11$ .

Example Program: Right Shift Operator(>>)

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{ 
    //Variable Declaration
     int A;
     
     // Variable initialization
     
     A = 23;
     
     //Bitwise Left Shift 
     
     A = A >> 1;
     
     // Printing the results
     
     cout << "Output = " << " " << A << endl;
     
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = 11

Complement(~)

The complement inverts the bits of a variable. If $A = 44$ with binary value $0010 \hspace{3px}1100$ then the 1’s complement binary value is $1101 \hspace{3px}0011$ which is $128 + 64 + 16 + 3 = 211$ .

The compiler gives the output in 2’s complement. To convert $1101 \hspace{3px}0011$ into 2’s complement.

\begin{aligned}&1 \hspace{5px}1 \hspace{5px}0\hspace{5px}1 \hspace{5px}0 \hspace{5px}0 \hspace{5px}1 \hspace{5px}1 \hspace{5px}\\ \\ &0 \hspace{5px}0 \hspace{5px}0\hspace{5px}0 \hspace{5px}0 \hspace{5px}0 \hspace{5px}0 \hspace{5px}1 \hspace{5px}\\ &-----------------------\\ &1 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}1 \hspace{5px}0 \hspace{5px}0\hspace{5px}\backslash \backslash Signed \hspace{5px} binary \hspace{5px}value\hspace{5px} of \hspace{5px}-45\\ &----------------------- \end{aligned}

Example Program: Complement

#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{ 
    //Variable Declaration
     short int A;
          
     // Variable initialization
     
     A = 45;
     
     //Bitwise Left Shift 
     
     A = ~A;
     
     // Printing the results
     
     cout << "Output = " << " " << A << endl;
     
    system("PAUSE");
    return EXIT_SUCCESS;
}

Output:

Output = -45