Get the prime factors of a number

Does your compiler not complain? If not, might want to turn warnings on. I just tried at repl.it and it rightly warns about control may reach end of non-void function. It still ran, but when I entered 2, I got the output 32690. And for 8, I got 2 2 0. So no, it doesn't run properly.

An unconditional return in a while loop doesn't make sense. I guess you meant if?

Let's say you found a large prime factor. The recursive call restarts the search at 2. Not efficient. You already know that none of the primes smaller than that large one divide the remaining n.

I'm going to address the coding, asking if an algorithm is a best approach is somewhat opinion based and to be avoided.

Avoid using abbreviations such as no. in prompts, it is better to use real words.

Iterative approaches may be better than recursive approaches because they use less resources (less memory for instance on the stack).

There is no reason to have a function prototype for the function primefactors if you swap the order of the functions primefactors() and main(). In C it is very common for the last function in a file to be main().

The indentation of the last } is problematic and should be corrected.

Leave horizontal space between operators such as , and & in all statements. that makes the code much more readable and easier to maintain. Examples:

 scanf("%d", &num); printf("%d", primefactors(num)); 

and

 for(int i=2; i <= sqrt(n); i++){ while(n % i == 0){ printf("%d ", i); return primefactors(n / i); } } if(n > 2) return(n); 

Leave vertical spacing between functions and sometimes separating logical blocks of code.

#include<stdio.h> #include<math.h> int primefactors(int n){ for(int i = 2; i <= sqrt(n); i++){ while(n % i == 0){ printf("%d ", i); return primefactors(n / i); } } if(n > 2) return(n); } int main() { int num; printf("Enter a no:-\n"); scanf("%d", &num); printf("%d", primefactors(num)); return 0; } 

Update with testing:

I have altered the code in the following way to get the prime factors for 2 through 100, the code fails on any power of 2, returning an unknown value for the final power of 2. It works properly in all other cases. This is approximately a 7% failure rate, an edge case fails.

#include<stdio.h> #include<math.h> int primefactors(int n) { for (int i = 2; i <= sqrt(n); i++) { while (n % i == 0) { printf(" %d ", i); return primefactors(n / i); } } if (n > 2) return(n); } int main() { for (int i = 2; i <= 100; i++) { printf("primefactors(%d) = ", i); printf("%d ", primefactors(i)); printf("\n"); } return 0; } 

"My solution runs properly", but not for all int

• primefactors(n_less_than_2) does not return anything. This hints that not all warnings were enabled. Save time and enable all compiler warnings.

• With primefactors(some_negative_value), code attempts the sqrt() of a negative leading to math woes.

• sqrt(some_perfect_square) is likely to form the exact correct square root, yet weak libraries exist and will instead return value just a little less, breaking the algorithm. Best to avoid floating point math for an integer problem.

• Should int unusually have more precision than double, then i<=sqrt(n) will not always form the correct test limit.

• Use i <= n/i instead. As it has a nearby n%i, a good compiler will likely form both the quotient and remainder for the time cost of one.

Incorporating the above fixes:

int primefactors(int n) { for (int i = 2; i <= n/i; i++) { while(n%i == 0){ printf("%d ", i); return primefactors(n/i); } } if (n>2) { return n; } return 1; // Unclear what OP really wants here. } 

Never ignore the return value from scanf():

scanf("%d",&num); printf("%d",primefactors(num)); 

Here, we have no idea whether num was assigned or not, so passing it to primefactors() is a bad idea.

Suggested replacement:

if (scanf("%d", &num) != 1) { puts("Input was not a valid integer\n", stderr); return EXIT_FAILURE; } printf("%d\n", primefactors(num)); 

(I also modified the printf() call to produce a complete line of output).

Having said that, it seems inappropriate to be reading from standard input at all - why not simply pass numbers as command-line arguments? Like this:

#include <limits.h> #include <stdio.h> #include <stdlib.h> int main(int argc, char*argv[]) { if (argc < 2) { fprintf(stderr, "Usage: %s NUMBER...\n", argv[0]); return EXIT_FAILURE; } while (*++argv) { char *end; long n = strtol(*argv, &end, 0); if (end == *argv || *end != '\0') { fprintf(stderr, "%s is not a valid number!\n", *argv); return EXIT_FAILURE; } if (n < INT_MIN || n > INT_MAX) { /* Why are we using signed numbers anyway? */ fprintf(stderr, "%s is outside the accepted range!\n", *argv); return EXIT_FAILURE; } printf("%d\n", primefactors((int)n)); } return EXIT_SUCCESS; /* optional for main() */ }