Function to Determine if Int is Prime

**Adrian** · 11-21-2007

I am supposed to write a function that determines if a number is prime or not. I return 1 if it is prime and 0 if it is not.

This is what I have so far:

Code:

int isPrime ( int number )
{
	int i = 2;

	for(i=2; i<number ; i++)
	{
		if((number&#37;i) == 0)
		{
		return 0;
		}
		else
		{
			return 1;
		}
         }
}

I'm not quite sure what's wrong with that.

Thanks

Edit: The function does work but I get the compiler warning, "control reaches end of non-void function" how could I get rid of that.

Edit Again: I figured it out, I just added an integer x and in one if statement I did x=0 rather than return 0 and for the other I did x = 1 rather than return 1, then at the end of the function I had return x.

Thanks anyways.

**brewbuck** · 11-21-2007

Originally Posted by Adrian

Edit Again: I figured it out, I just added an integer x and in one if statement I did x=0 rather than return 0 and for the other I did x = 1 rather than return 1, then at the end of the function I had return x.

Your real problem is that you "return 1;" when there is a non-zero remainder. That's obviously not right, you need to check ALL the values. Instead of returning 1, just don't do anything at all. And at the end of the function, outside the for-loop, return 1.

You have effectively accomplished the same thing by introducing this "x" variable but it's not necessary, and still wrong. Once x has been set to 0, it must never be set back to 1, since you've already proven the number isn't prime. And there is no point continuing the loop once this happens.

What does your current code look like?

**Adrian** · 11-21-2007

Code:

int isPrime ( int number )
{

	int x;

	int i = 2;

	for(i=2; i<number ; i++)
	{
		if((number&#37;i) == 0)
		{
		x = 0;
		}
		else
		{
		x = 1;
		}
	}

	return x;
}

But if there is a remainder that is 0 when the number is divided by anything between 1 and itself doesn't that mean it's not prime?

**Adrian** · 11-21-2007

This is what it should be:

Code:

int isPrime ( int number )
{

	int x = 1;

	int i = 2;

	for(i=2; i<number ; i++)
	{
		if((number&#37;i) == 0)
		{
		x = 0;
		}

	}


	return x;
}

That else statement in the for loop was messing things up.

I'm going to go now, but thanks for your help, the function does work now.

Thanks!

**Dino** · 11-22-2007

Here's a tip.

Let's say the number passed to your program is 29 - a primer number, but we don't know that yet. The current logic will test all numbers between 2 and 28. That's wasteful, because 29/2 (or generically, 1/2 of the passed number) is the largest value you ever need to check. Run some other samples through, you'll see what I mean.

Todd

**IceDane** · 11-22-2007

Originally Posted by Todd Burch

Here's a tip.

Let's say the number passed to your program is 29 - a primer number, but we don't know that yet. The current logic will test all numbers between 2 and 28. That's wasteful, because 29/2 (or generically, 1/2 of the passed number) is the largest value you ever need to check. Run some other samples through, you'll see what I mean.

Todd

Isn't it the square root of the max number the largest number you have to check?

**Dino** · 11-22-2007

You could do that too.

**Salem** · 11-22-2007

> I return 1 if it is prime and 0 if it is not.
So you return 1 ONLY if all the tests pass, not the first test.
Your first attempt was isOdd(), not isPrime().

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