Is a Recursive-Iterative Method Better than a Purely Iterative Method to find out if a number is prime?

I made this program in C that tests if a number is prime. I'm as yet unfamiliar with Algorithm complexity and all that Big O stuff, so I'm unsure if my approach, which is a combination of iteration and recursion, is actually more efficient than using a purely iterative method.

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

typedef struct primenode{
    long int key;
    struct primenode * next;
}primenode;

typedef struct{
    primenode * head;
    primenode * tail;
    primenode * curr;
    unsigned long int size;
}primelist;

int isPrime(long int number, primelist * list ,long int * calls, long int * searchcalls);
primenode * primelist_insert(long int prime, primelist * list);
int primelist_search(long int searchval, primenode * searchat, long int * calls);
void primelist_destroy(primenode * destroyat);

int main(){
    long int n;
    long int callstoisprime = 0;
    long int callstosearch = 0;
    int result = 0;
    primelist primes;

    //Initialize primelist
    primes.head = NULL;
    primes.tail = NULL;
    primes.size = 0;

    //Insert 2 as a default prime (optional step)
    primelist_insert(2, &primes);

    printf("\n\nPlease enter a number: ");
    scanf("%d",&n);
    printf("Please wait while I crunch the numbers...");
    result = isPrime(n, &primes, &callstoisprime, &callstosearch);
    switch(result){
        case 1: printf("\n%ld is a prime.",n); break;
        case -1: printf("\n%ld is a special case. It's neither prime nor composite.",n); break;
        default: printf("\n%ld is composite.",n); break;
    }
    printf("\n\n%d calls made to function: isPrime()",callstoisprime);
    printf("\n%d calls made to function: primelist_search()",callstosearch);

    //Print all prime numbers in the linked list
    printf("\n\nHere are all the prime numbers in the linked list:\n\n");
    primes.curr = primes.head;
    while(primes.curr != NULL){
        printf("%ld ", primes.curr->key);
        primes.curr = primes.curr->next;
    }
    printf("\n\nNote: Only primes up to the square root of your number are listed.\n"
                "If your number is negative, only the smallest prime will be listed.\n"
                "If your number is a prime, it will itself be listed.\n\n");

    //Free up linked list before exiting
    primelist_destroy(primes.head);

    return 0;
}

int isPrime(long int number, primelist * list ,long int * calls, long int *searchcalls){
//Returns 1 if prime
//          0 if composite
//          -1 if special case
    *calls += 1;
    long int i = 2;
    if(number==0||number==1){
        return -1;
    }
    if(number<0){
        return 0;
    }
    //Search for it in the linked list of previously found primes
    if(primelist_search(number, list->head, searchcalls) == 1){
        return 1;
    }
    //Go through all possible prime factors up to its square root
    for(i = 2; i <= sqrt(number); i++){ 
        if(isPrime(i, list,calls,searchcalls)){
            if(number%i==0) return 0; //It's not a prime
        }
    }
    primelist_insert(number, list); /*Insert into linked list so it doesn't have to keep checking
                                                if this number is prime every time*/
    return 1;
}

primenode * primelist_insert(long int prime, primelist * list){
    list->curr = malloc(sizeof(primenode));
    list->curr->next = NULL;

    if(list->head == NULL){
        list->head = list->curr;
    }
    else{
        list->tail->next = list->curr;
    }
    list->tail = list->curr;
    list->curr->key = prime;
    list->size += 1;

    return list->curr;
}

int primelist_search(long int searchval, primenode * searchat, long int * calls){
    *calls += 1;
    if(searchat == NULL) return 0;
    if(searchat->key == searchval) return 1;
    return primelist_search(searchval, searchat->next, calls);
}

void primelist_destroy(primenode * destroyat){
    if(destroyat == NULL) return;
    primelist_destroy(destroyat->next);
    free(destroyat);
    return;
}

Basically, a lot of what I've seen simple primalty tests do is: 0. 2 is a prime. 1. Cycle through all integers from 2 to half or the square root of the number being tested. 2. If the number is divisible by anything, break and return false; it's composite. 3. Otherwise, return true after the last iteration; it's prime.

I figured that you don't have to test against every number from 2 to the square root, just every prime number, because all other numbers are multiples of primes. So, the function calls itself to find out if a number is prime before using the modulus on it. This works, but I thought it a bit tedious to keep testing all those primes over and over again. So, I used a linked list to store every prime found in it as well, so that before testing primalty, the program searches the list first.

Is it really faster, or more efficient, or did I just waste a lot of time? I did test it on my computer, and for the larger primes it did seem faster, but I'm not sure. I also don't know if it uses significantly more memory since Task Manager just stays a constant 0.7 MB whatever I do.

Thanks for any answers!

Since your program tests just one number, you're wasting time trying to avoid testing by composites. You perform a lot of computations to save one meager modulo operation.

If you were testing more than a few numbers in a row for primality, then it would make sense to pre-compute the primes up to a sqrt of the top limit of that range, and go through those primes when testing the candidates.

Better yet will be to perform an offset sieve of Eratosthenes (C code here), to find the primes in a given range. The time complexity of the sieve of Eratosthenes to find primes from 2 to N is O(N log log N); and trial division by primes up to the sqrt, O(N^1.5 / (log N)^2) (which is worse; e.g. the ratio of run times for the sieve up to 1mln compared to 100k is 10.7x, vs 22x for trial division; 2mln vs 1 mln is 2.04x for the sieve, 2.7x for the trial division).

Pseudocode for the offset sieve of Eratosthenes:

Input: two Integers n >= m > 1

Let k = Floor(Sqrt(n)),
Let A be an array of Boolean values, indexed by Integers 2 to k, and
    B an array of Booleans indexed by Integers from m to n,
    initially all set to True.

for i = 2, 3, 4, ..., not exceeding k:
  if A[i] is True:
    for j = i^2, i^2+i, i^2+2i, ..., not greater than k:
      A[j] := False
    for j = i^2, i^2+i, i^2+2i, ..., between m and n, inclusive:
      B[j] := False

Output: all `i`s such that B[i] is True, are all the primes 
                                     between m and n, inclusive.

A common optimization is to work with odds only, i = 3,5,7,..., avoiding any even numbers from the outset (2 is known to be a prime anyway, and any even number is a composite). Then the step of 2i, and not just i, can be used in both inner loops. Thus the even indices are excluded from processing altogether (usually with condensed addressing schemes, val = start + 2*i).