Prime or not?

[RobGillespie]

So I've finished a code but when I submitted it to the grading program, it says this:
cc1plus: warnings being treated as errors
Primenum.cpp: In function 'bool isPrime(int)':
Primenum.cpp:17: error: control reaches end of non-void function
compilation terminated due to -Wfatal-errors.

I've had this before and basically it's saying that my code eventually runs to a point where my isPrime(int) function doesn't return anything. I've solved this issue in the past but it was the beginning of the semester and I don't remember what I had to do. Here is my code:

#include <iostream>
using namespace std;

bool isPrime(int a)
{
	int prime;
	for( int i = 2; i < a; i++ )
	{
		prime = a % i;
		if( prime == 0)
		{
			return false;
		}
	}

	if(prime != 0)
		return true;
}

int main()
{
	float number;
	cout << "Enter number: ";
	cin >> number;

	cout << endl;

	if( isPrime(number) == true)
	{
		cout << number << " is prime.";
	}
	else
	{
		cout << number << " is not prime.";
	}
}

I get all the correct answers unless I type in 2 or lower, then the answer is wrong. I'm still working on that as well so if someone knows what's going on there that would be good to know. I don't know why 2 is considered prime in my code because if 2 was entered: 2 % 2 = 0 so when 0 is put in place for "prime", it is supposed to return false which means it's not prime.

[Zhuge]

To explain the compiler error:
Suppose you reach line 16 and prime is 0. What does the function return then?

Why 2 and below are prime by your code:
a is <2, so and i=2 is not less than a<=2, so the loop is not entered. prime, being uninitialized, is likely to not be zero, so the function returns true.

[RobGillespie]

At line 16, if prime is 0, then it would have already returned false from the for loop would it not?

I simply inserted this:

if(a == 2)
	{
		return false;
	}

To return "not prime" if 2 was entered.

[Smac89]

bool isPrime(int a)
{
        if (~a & 1 && a != 2)
               return false;

	int prime;
	for( int i = 2; i < a; i++ )
	{
		prime = a % i;
		if( prime == 0)
		{
			return false;
		}
	}

	return true;
}

[RobGillespie]

I've never seen this before:

   if (~a & 1) // if a is even...
            return true;

What does that mean?
Keep in mind I'm very new to programming.

[Smac89]

It is equivalent to (a % 2) == 0. It is the bitwise version of that

[RobGillespie]

Ok. I see. So really I needed the "return true" at the very end of the function.

[RobGillespie]

My compile program is now saying that 2 has to be prime so I can just forget about the part that makes 2 not prime...even though 2 is not prime.

[Zhuge]

Er, 2 is a prime number. Could you try explaining the issue a different way, maybe...?

[RobGillespie]

Oh wow it is a prime number...I'm an idiot. I was thinking too much into modular division...

The answer to my problem was that I had no return value if the other conditions weren't met like you said.

Thanks for the help everyone.

[Hiten]

Yeap, 2 is a prime number. And in your loop you could test just until i <=sqrt(a).
If there is no i < sqrt(a) for which ( a%i )== 0 then there won't be an i between sqrt(a)and a for which ( a%i )== 0 as well.