Given a floating point number, write a function to count set bits in its binary representation.
For example, floating point representation of 0.15625 has 6 set bits (See this). A typical C compiler uses single precision floating point format.
We can use the idea discussed here. The idea is to take address of the given floating point number in a pointer variable, typecast the pointer to char * type and process individual bytes one by one. We can easily count set bits in a char using the techniques discussed here.
Following is C implementation of the above idea.
#include <stdio.h> // A utility function to count set bits in a char. // Refer http://goo.gl/eHF6Y8 for details of this function. unsigned int countSetBitsChar( char n)
{ unsigned int count = 0;
while (n)
{
n &= (n-1);
count++;
}
return count;
} // Returns set bits in binary representation of x unsigned int countSetBitsFloat( float x)
{ // Count number of chars (or bytes) in binary representation of float
unsigned int n = sizeof ( float )/ sizeof ( char );
// typecast address of x to a char pointer
char *ptr = ( char *)&x;
int count = 0; // To store the result
for ( int i = 0; i < n; i++)
{
count += countSetBitsChar(*ptr);
ptr++;
}
return count;
} // Driver program to test above function int main()
{ float x = 0.15625;
printf ( "Binary representation of %f has %u set bits " , x,
countSetBitsFloat(x));
return 0;
} |
Output:
Binary representation of 0.156250 has 6 set bits
Time Complexity: O(nlogn)
Auxiliary Space: O(1)