Function return value in C [closed] - c

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I am an absolute newbie to the C language (I started today to learn it). As a test I wrote this program to output all prime numbers between two ranges. I then optimized and shortened the checkIfPrime function with some google help and took all return statements out of the function, for it obviously doesn't need to have a return value to understand I want to have the flag variable back. How does the function returns the flag variable even without any returns in it?
#include <stdio.h>
int checkIfPrime(int n);
int main(int argc, char *argv[]) {
int n1, n2;
scanf("%d %d", &n1, &n2);
for (int i = n1; i < n2; i++) {
if (checkIfPrime(i)) {
printf("%s", "Is prime: ");
printf("%d\n", i);
}
}
return 0;
}
int checkIfPrime(int n) {
int flag = 1;
for (int i = 2; i <= (n / 2); i++) {
if (n % i == 0) {
flag = 0;
break;
}
}
}

Answer 1

Unlike some languages such as Groovy where the last expression evaluated by a function is the return value of the function, in C you need to explicitly state the return value with a return statement.
int checkIfPrime(int n)
{
int flag = 1;
for (int i = 2; i <= (n / 2); i++) {
if (n % i == 0) {
flag = 0;
break;
}
}
return flag;
}
If you fail to explicitly return a value from a function that expects one, your code invokes undefined behavior.
As an example of undefined behavior, your original code give the "correct" answer for me, but if I change the function to this:
int checkIfPrime(int n) {
int flag = 1;
for (int i = 2; i <= (n / 2); i++) {
if (n % i == 0) {
flag = 0;
break;
}
}
// this loop should have no effect
for (int i = 2; i <= (n / 2); i++) {
}
}
In a properly behaving program, the extra loop added should have no effect. But without a return, I get back this if I input 2 and 10:
Is prime: 2
Is prime: 3
Is prime: 4
Is prime: 5
Is prime: 6
Is prime: 7
Is prime: 8
Is prime: 9
If you had put the return statement in place at the end of the function, things like this wouldn't happen.
Why the above code behaves the way it does is an implementation detail of the compiler in use. A different compiler may have different output, or the same compiler with different optimization settings could have different output.

Answer 2

Your current function int checkIfPrime(int n) is by its definition saying it will return an int, so you must then return an int value.
If you really didn't want to return a value using the return statement, you could pass the value back as a pointer to a variable to store the flag in:
void checkIfPrime(int n, int *flag) {
*flag = 1;
for (int i = 2; i <= (n / 2); i++) {
if (n % i == 0) {
*flag = 0;
break;
}
}
}
Its far easier to read and understand using a return of the value though, and main would then have to change to allocate a new variable and would have to call the function and then check the variable - much nicer just using the return value.

Answer 3

There are only two ways to give a value back from a function: explicitly return a value, or mutate a variable external to the function.
Except when writing OO code (and in that case you're mutating object fields), or a void function, your functions should always explicitly return a value.
Since you gave your function a return type, I'm surprised the compiler didn't raise an error.
You shouldn't try to avoid using return statements; avoid using them in ways that lead to confusing code.

Answer 4

How does the function returns the flag variable even without any returns in it?
It doesn't; or, it doesn't do so in a predictable or reliable manner.
The intent is that if you define a function T f() { ... } and T is not void, then you will have at least one return statement returning a value of type T.
For whatever reason, the language definition does not mandate a return statement in a non-void function1 (the compiler may complain, but it's not required to), but if the caller attempts to use the return value of the function and you don't explicitly return a value, then the behavior is undefined, meaning any result you get (whether expected or not) is considered "correct".
Interestingly, it does mandate that a return statement in a void function not return a value.