Trying to fill an array with a binary value and determine least significant bit

Question

So I have the task of creating a program that will take exactly two arguments, the second being an integer. I have to convert that integer to a binary value and then print out the location of the least significant "1" bit. Below is the code I currently have that successfully converts the integer in to binary. I wanted to fill an array with the binary value and then cycle from the last position and decrement a counter until I encountered a "1" and then print out the position of that bit by subtracting the overall length of the array from the current position in the array.

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

int convert(int);
int main(int argc, char *argv[])
{
    int  bin;
    int n = atoi(argv[2]);

        if (argc<2)
        {
                printf("\nNot Enough Arguments.");
        }
        if (argc>2)
        {
                printf("\nToo Many Arguments.");
        }
        if (argc==2)
        {
                scanf("%d", &n);
                bin = convert(n);
                lessbit[] = bin;
        } 

    return 0;
}

int convert(int n)
{
    if (n == 0)
    {
        return 0;
    }
    else
    {
        return (n % 2 + 10 * convert(n / 2));
    }
} 

Answer 1

Numbers are stored in the machine as binary so you do not need to convert anything.

Here is a small demo program that will print the position of the first 1 moving from right to left (least significant 1 bit).

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

int main(int arc, char* argv[]) {
    if(argc != 2)
        exit(EXIT_FAILURE);

    int bin = atoi(argv[1]);
    unsigned counter = 0;

    while(bin % 2 == 0) {
        bin >>= 1;
        counter++;
    }

    printf("%u\n", counter);
    return 0;
}

This code prints out the exponent base 2. So if the number is 8 which is 1000 in binary the output will be the position of the 1 which is 3 (since it's in the 2^3 position).