number is in power series of 2

Write a Program to Find Given Number is in power series of 2 or Not?

Logic 1:
If Number Will keep on divide by 2 till the number become 1

int find_in_power_serise(int number, int power){
        do{
                if((number % power) ==0){
                        number = number/power;
                }else{
                        break;
                }
        }while(number > 1);
        if(number == 1){
                return 1;
        }else{
                return 0;
        }
}

Logic 2:
If we will see binary re-presentation of a Number.
we can come to know that if a single bit set then that number is in power series of 2.

int is_nth_bit_set(unsigned int num, int bit){
        return ((num >> bit) & 0x01);
}

int find_in_power_serise_of_2(unsigned int number){
        int count = 0;
        for(int i=0; i < (sizeof(number)*8) ; i++){
                if(is_nth_bit_set(number, i)){
                        count ++;
                }
        }
        if(count == 1){
                return 1;
        }else{
                return 0;
        }
}

without Loop and Without using % operator

int count_bit_set(unsigned int number, int count)
{
        if(number){
                if(number&0x01)
                        count++;
                return count_bit_set((number >> 1), count);
        }
        return count;
}

int find_in_power_serise_of_2_no_loop(unsigned int number){
        if(count_bit_set(number, 0)==1){
                return 1;
        }else{
                return 0;
        }
}

full code

#include <stdio.h>
#include <stdlib.h>

int find_in_power_serise(int number, int power){
        do{
                if((number % power) ==0){
                        number = number/power;
                }else{
                        break;
                }
        }while(number > 1);
        if(number == 1){
                return 1;
        }else{
                return 0;
        }
}

int is_nth_bit_set(unsigned int num, int bit){
        return ((num >> bit) & 0x01);
}

int find_in_power_serise_of_2(unsigned int number){
        int count = 0;
        for(int i=0; i < (sizeof(number)*8) ; i++){
                if(is_nth_bit_set(number, i)){
                        count ++;
                }
        }
        if(count == 1){
                return 1;
        }else{
                return 0;
        }
}

int count_bit_set(unsigned int number, int count)
{
        if(number){
                if(number&0x01)
                        count++;
                return count_bit_set((number >> 1), count);
        }
        return count;
}

int find_in_power_serise_of_2_no_loop(unsigned int number){
        if(count_bit_set(number, 0)==1){
                return 1;
        }else{
                return 0;
        }
}

int main(int argc, char *argv[]){
        if(argc != 2){
                fprintf(stderr, "USAGE:%s <number>\n", argv[0]);
                return -1;
        }
        int num = atoi(argv[1]);
        int power = 2;
        printf("%d is %s Power Serise of %d\n", num, find_in_power_serise(num, power)?"IN":"NOT IN", power);
        printf("%d is %s Power Serise of %d\n", num, find_in_power_serise_of_2(num)?"IN":"NOT IN", 2);
        printf("%d is %s Power Serise of %d\n", num, find_in_power_serise_of_2_no_loop(num)?"IN":"NOT IN", 2);
        return 0;
}

Compilation and Output

rajesh@ideapad:~/Rajesh/Blog/math$ gcc power_serise_2.c
rajesh@ideapad:~/Rajesh/Blog/math$ ./a.out 64
64 is IN Power Serise of 2
64 is IN Power Serise of 2
64 is IN Power Serise of 2
rajesh@ideapad:~/Rajesh/Blog/math$ ./a.out 65
65 is NOT IN Power Serise of 2
65 is NOT IN Power Serise of 2
65 is NOT IN Power Serise of 2