C - 尝试将指向函数的指针作为参数传递时出错，无法获取值

问：

我无法将指向函数的指针作为参数传递，同时能够成功显示指向的值。函数 checkIfPrime 将不起作用，也不会接收指针 pAr 指向的值。对不起，代码混乱。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <malloc.h>

int calculatePrime(){
    int i = 0, j = 0, z = 0, check = 0;
    int prime[100] = {2, 3, 5};
    int size = 3, primeNum = 7;
    int *pAr = prime;

    while (i <= 100){
        for (z = 0; z <= size - 1; z++){
            if (primeNum % prime[z] != 0){
                check++;
            }
        }
        if (check == size){
            prime[size] = primeNum;
            size++;
        }
        i++;
        primeNum += 2;
        check = 0;
    }
    return pAr;
}

int checkIfPrime(int *pAr, int num){
    int i, check = 0, isPrime = 0;
    int fail = 0;

    printf("\n\n");

    printf("Doesn't works here, *(pAr + 5): %d", *(pAr + 5));

    for(i = 0;i <= 45 && fail == 0; i++){
        if (num % *(pAr + i) != 0){
            check++;
        }
        else{
            fail = 1;
        }
    }

    if (check == 45){
        isPrime = 1;
        printf("\n%d is prime\n", i);
    } else{
        isPrime = 0;
        printf("\n%d is not prime\n", i);
    }

    return isPrime;
}

int main(){
    FILE *fp = NULL;
    fp = fopen("test.txt", "r");

    int i = 0, j = 0, k = 0;
    int even[20], odd[20], prime[20];

    int num, temp;
    char str[20];
    int isPrime;
    int ar[45];
    int *pAr;

    pAr = calculatePrime();
    printf("Works here, *(pAr + 5): %d\n", *(pAr + 5));

    isPrime = checkIfPrime(&pAr, 7);
    //Having issues here

    return 0;
}