incompatible pointer compiler error - c

Question

I have this assignment where we expected to write a program that takes
a positive integer number as a command line argument and prints the
smallest prime number bigger than the given number.
The main function shouldn't be edited, however, you may create a
header file for defining needed functions.
So far this is what I came up with, I just can't find out what
is wrong with my program. Help is appreciated.
main function: I can't edit the main function, however, you can create a header
#include "slow-prime.h"
int main(int argc, char *argv[]) {
int num;
int nxt;enter code here
int ret = EXIT_FAILURE;
if (argc < 2) {
printf("error: missing command line argument\n");
goto ERROR;
if (get_number(argv[1], &num)) {
printf("error: %s not a number\n", argv[1]);
goto ERROR;
}
next_prime(num, &nxt);
printf("%d\n", nxt);
ret = EXIT_SUCCESS;
ERROR:
return ret;
}
}
needed functions are created in the slow-prime.h
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// define true and false
#define true 1
#define false 0
// check whether the numer is prime or mnot
int isPrime(int num){
if (num < 2) {
return false;
}
for (int i = 2; i <= num / i; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
// get number
void get_number(char *argv[], int num) {
num = atoi(argv[1]);
}
// loop through the numbers/ and pick the one
void next_prime(int num, int next){
for(int i = 2; i < 80; i++){
if (isPrime(i) == true){
next = i;
if (next > num) {
return exit(0);
}
}
}
}
Error Message:
error

Answer 1

You have:
void get_number(char *argv[], int num) {
and you are calling the function using:
if (get_number(argv[1], &num)) {
You are passing the wrong types for both arguments.
argv[1] is of type char*.
&num is of type int*.
I think you should to use:
void get_number(char *arg, int *num) {
and change the implementation slightly.
void get_number(char *arg, int *num) {
*num = atoi(arg);
}
Also, given that get_number returns void, you cannot use it in the conditional of an if statement. You'll need to change its return type to something else, an int perhaps. In that case, using atoi might not be appropriate. atoi returns 0 when it cannot convert the string to an int. If 0 a valid value for you, then atoi is not a good choice. However, you can use sprintffor all cases.
If 0 is not a valid number, you can use:
int get_number(char *arg, int *num) {
*num = atoi(arg);
return *num;
}
If 0 is a valid number, you can use:
int get_number(char *arg, int *num) {
return (sprintf(arg, "%d", num) == 1);
}