How to extract ‘k’ bits from a given position ‘p’ in a number?
Examples:
Input : number = 171 k = 5 p = 2 Output : The extracted number is 21 171 is represented as 10101011 in binary, so, you should get only 10101 i.e. 21. Input : number = 72 k = 5 p = 1 Output : The extracted number is 8 72 is represented as 1001000 in binary, so, you should get only 01000 i.e 8.
1) Right shift number by p-1.
2) Do bit wise AND of k set bits with the modified number. We can get k set bits by doing (1 << k) – 1.
C++
// C++ program to extract k bits from a given // position. #include <bits/stdc++.h> using namespace std;
// Function to extract k bits from p position // and returns the extracted value as integer int bitExtracted( int number, int k, int p)
{ return (((1 << k) - 1) & (number >> (p - 1)));
} // Driver code int main()
{ int number = 171, k = 5, p = 2;
cout << "The extracted number is " <<
bitExtracted(number, k, p);
return 0;
} // This code is contributed by importantly |
C
// C program to extract k bits from a given // position. #include <stdio.h> // Function to extract k bits from p position // and returns the extracted value as integer int bitExtracted( int number, int k, int p)
{ return (((1 << k) - 1) & (number >> (p - 1)));
} // Driver code int main()
{ int number = 171, k = 5, p = 2;
printf ( "The extracted number is %d" ,
bitExtracted(number, k, p));
return 0;
} |
Java
// Java program to extract k bits from a given // position. class GFG {
// Function to extract k bits from p position
// and returns the extracted value as integer
static int bitExtracted( int number, int k, int p)
{
return ((( 1 << k) - 1 ) & (number >> (p - 1 )));
}
// Driver code
public static void main (String[] args) {
int number = 171 , k = 5 , p = 2 ;
System.out.println( "The extracted number is " +
bitExtracted(number, k, p));
}
} |
Python3
# Python program to extract k bits from a given # position. # Function to extract k bits from p position # and returns the extracted value as integer def bitExtracted(number, k, p):
return ( (( 1 << k) - 1 ) & (number >> (p - 1 ) ) );
# number is from where 'k' bits are extracted # from p position number = 171
k = 5
p = 2
print ( "The extracted number is " , bitExtracted(number, k, p))
|
C#
// C# program to extract k bits from a given // position. using System;
class GFG {
// Function to extract k bits from p position
// and returns the extracted value as integer
static int bitExtracted( int number, int k, int p)
{
return (((1 << k) - 1) & (number >> (p - 1)));
}
// Driver code
public static void Main()
{
int number = 171, k = 5, p = 2;
Console.WriteLine( "The extracted number is "
+ bitExtracted(number, k, p));
}
} //This code is contributed by Anant Agarwal. |
PHP
<?php //PHP program to extract // k bits from a given // position. // Function to extract k // bits from p position // and returns the extracted // value as integer function bitExtracted( $number , $k , $p )
{ return (((1 << $k ) - 1) &
( $number >> ( $p - 1)));
} // Driver Code
$number = 171; $k = 5; $p = 2;
echo "The extracted number is " ,
bitExtracted( $number , $k , $p );
// This code is contributed by Ajit ?> |
Javascript
<script> // JavaScript program to extract k bits from a given // position. // Function to extract k bits from p position // and returns the extracted value as integer function bitExtracted(number, k, p)
{ return (((1 << k) - 1) & (number >> (p - 1)));
} // Driver code let number = 171, k = 5, p = 2;
document.write( "The extracted number is " ,
bitExtracted(number, k, p));
// This code is contributed by Manoj. </script> |
Output :
The extracted number is 21
Time Complexity: O(1)
Auxiliary Space: O(1)
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