/* C program to accept a decimal number and convert it binary and count the number of 1′s in the binary number */

#include <stdio.h>

void main()
{
long num, dnum, rem, base = 1, bin = 0, no_of_1s = 0;

printf(“Enter a decimal integer\n”);
scanf(“%ld”, &num);

dnum = num;

while( num > 0 )
{
rem = num % 2;
if ( rem == 1 )      /*To count no.of 1s*/
{
no_of_1s++;
}
bin = bin + rem * base;
num = num / 2 ;
base = base * 10;
}
printf(“Input number is                   = %d\n”, dnum);
printf(“Its Binary equivalent is          = %ld\n”, bin);
printf(“No.of 1′s in the binary number is = %d\n”, no_of_1s);

}             /* End of main() */

/*————————————————–
Output
Enter a decimal integer
75
Input number is                   = 75
Its Binary equivalent is          = 1001011
No.of 1′s in the binary number is = 4

RUN2
Enter a decimal integer
128
Input number is                   = 128
Its Binary equivalent is          = 10000000
No.of 1′s in the binary number is = 1

—————————————————–*/

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Tagged as: binary number, C program to accept a decimal number and conert it binary and count the number of 1's in the binary number, count the number of 1's in the binary number