Breaking out of a loop without a break statement [C] - c

Question

How do you break out of a loop without a break statement? My professor HATES break statements and tells us not to use it. I'm just curious how would I break out of the while-loop if the number I got WAS NOT a prime number?
Here's my code:
#include <stdio.h>
#include <stdlib.h>
/* Prototypes */
void primeChecker(int num1);
int main() {
int num1 = 5;
primeChecker(num1);
return 0;
}
void primeChecker(int num1) {
int i, flag = 0;
printf("Enter a number to check for prime numbers: ");
scanf("%d", &num1);
/* Number to start with */
i = 2;
while (i <= num1/2) {
if (num1 % i == 0) {
flag = 1;
} else {
i++;
}
}
if (flag == 0) {
printf("The number is a prime number!");
} else {
printf("The number is NOT a prime number!");
}
}

Answer 1

Or
int prime = 1;
while (i <= num1/2 && prime) {
if (num1 % i == 0){
prime = 0;
} else {
i++;
}
}
if(prime){
printf("The number is a prime number!");
}else{
printf("The number is not prime.");
}
I mean, you almost had it:
while (i <= num1/2 && !flag){
would have done the trick as well

Answer 2

You can do
while (i <= num1/2) {
if (num1 % i == 0) {
i = num1;
} else {
i++;
}
}
This makes i larger then num1/2 and the while loop exits.
You probably need some more changes to make this work.

Answer 3

In your case, you can use the value of flag as condition:
while (flag == 0 && i <= num1/2) {
if (num1 % i == 0) {
flag = 1;
} else {
i++;
}
}
But that looks like Pascal rather than C. A better solution might be to refactor the loop so that it is in a separate function:
int is_prime(int num1)
{
int i = 2;
while (i <= num1/2) {
if (num1 % i == 0) return 0;
i++;
}
return 1;
}
This makes the code simpler and separates the input stuff in primeChecker from the actual prime checking.