Is this example of recursion correct? - c

Question

int main() {
int num;
printf("Enter a number\n");
scanf(" %d", &num);
num = prime(num);
if (num == 0)
printf("This is not a prime number");
if (num == 1)
printf("This is a prime number");
}
prime(int num) {
static i = 2;
if (num % i == 0)
return 0;
prime(i + 1);
return 1;
}
Note that it doesn't work in some compilers.
I want to know if we can call it recursion or not.
Specifically, I am confused if calling prime function like prime(i + 1) falls under recursion or not.

Answer 1

In your code your were passing i as argument in further recursive calls which was logically not correct. You need to pass the number as well as a divisor the function prime. The below code is one of the way to implement your logic.
#include<stdio.h>
int prime(int num,int i)
{
if(num==i)
return 1;
if(num % i == 0)
return 0;
return prime(num,i+1);
}
int main()
{
int num;
printf("Enter a number\n");
scanf(" %d", &num);
if(num>1){
num = prime(num,2);
if(num == 0)
printf("This is not a prime number");
else
printf("This is a prime number");
}
else printf("This is neither prime nor composite");
return 0;
}
Note the no input validation is performed and it is assumed that user will input only positive integers.

Answer 2

You are calling prime with i during the recursive call, which is not the actual value you want to check for prime, that is the reason it is not working!
It will return 0 for multiples of 2 only otherwise it will be stuck in an infinite loop.
int prime(int num) {
static i = 2;
if(num % i == 0)
return 0;
prime(i + 1); // passing i instead of num
return 1;
}
You can pass the divisor along with the number, and increase the divisor with 1 in every recursive call.
int prime(int num, int d) {
if(num == 0 || num == 1 || num % d == 0)
return 0;
if(d <= sqrt(num)) // num will not be divisible by d > sqrt(num)
prime(num, d + 1);
return 1;
}
Initially call prime with divisor 2 like prime(num, 2) in main.