Prime factors of a number without using a flag variable? - c

Question

I tried finding prime factors of a number without using a flag variable but I don't know what's the fault in the code.
#include <stdio.h>
int main(int argc, char const *argv[])
{
int num, i, j;
printf("Enter a number: ");
scanf("%d", &num);
for (i = 2; i <= num; i++)
{
if (num % i == 0)
{
for (j = 2; j <= i / 2; j++)
{
if (i % j == 0)
{
printf("%d", i);
}
}
}
}
return 0;
}

Answer 1

You have to print values that are factor of num and prime.
In your code you are printing factors of num, but not prime by printing i after seeing that a j that divids i exists.
Instead of that, you should check for all candidates of j and print i after seeing all js tested don't divide i.
Also you may want to add some delimiters between the factors to print.
Try this:
#include <stdio.h>
int main(int argc, char const *argv[])
{
int num, i, j;
printf("Enter a number: ");
scanf("%d", &num);
for (i = 2; i <= num; i++)
{
if (num % i == 0)
{
for (j = 2; j <= i / 2; j++)
{
if (i % j == 0)
{
break; /* stop iteration because a factor is found */
}
}
if (j > i / 2) /* check if the iteration ended not by finding factor but by checking all candidates */
{
printf("%d\n", i);
}
}
}
return 0;
}

Answer 2

Your approach feels a bit convoluted. Perhaps just do something like:
#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char **argv)
{
unsigned num = argc > 1 ? strtoul(argv[1], NULL, 10) : 137;
int count = 0;
printf("%d = ", num);
for( unsigned i = 2; i <= num; i++ ){
while( num % i == 0 ){
printf("%s %d", count++ ? " *" : "", i);
num /= i;
}
}
printf(" ( %d factor%s )\n", count, count == 1 ? "" : "s");
return 0;
}