Wsrtjtyk

I want to list all the factors of a number into an array. So, here's my try at it.

#include<stdio.h>

#define n1 10

int main()

{

     int n,i,j,a[n1];

     printf("Enter num: ");

     scanf("%d",&num);

     for(i=1;i<n;i++){

        if(n%i==0){

           for(j=0;j<n;j++)

I don't know how to proceed from here. I'm confused.

#include<stdio.h>

#define n1 10

int main()

{

        int n,i,j,a[n1];

        printf("Enter number : ");

        scanf("%d",&n);

        j = 0;

        for(i=1;i<n;i++){

                if(n%i==0)

                        a[j++] = i;

        }

        printf("The factors of the number are: ");

        for(j=0;j;j++)

                printf("%dn",a[j]);

}

You are quite close, but the inner loop seems pointless. Once you've found a factor, just store it and keep track of how many you've found:

 int factors(int n, int *factors)

 {

  int j = 0;

  for (int i = 2; i < n; ++i)

  {

    if (n % i == 0)

    {

      factors[j++] = i;

    }

  } 

  return j;

}

I ran this on 471113, and got [193, 2441] which checks out.

Do note that this is quite limited by the max precision of int (and should use unsigned long and be safer for real code).

I think what you are looking for is:

for(i=0, j=0;i<n;i++)

    if(n%i==0)

        a[j++] = i;



// print it

for (i=0; i<j; i++)

    printf("%d ",a[i]);