My loop is not so infinite?

the original way will not accurately determine if a number is prime, it will only determine if a number is odd, not all odd numbers are prime ( 9 and 15 for example). The method I posted will absolutely determine if its prime or not. Although the method I gave you is probably the slowest 'fast' method there is, its also the simplest. There are far faster methods otu there, but the maths gets thick.

Quote:

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Originally Posted by Caduceus

Aha! I never thought of using logical-or! I tried to implement a break if the user enters zero, but that didn't work :P Sorry to keep asking questions, but I've read the short tutorial, and I can't figure it out, how does break work?

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Though you may not need break here, a break keyword simple aborts a loop, the loop it is written in:

Code:

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int main()
{
        for (int i = 0; i < 100000; i++)
        {
                if (i == 100) break;
        }
        return 0;
}

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When i hits 100, the loop aborts due to the break keyword.

Quote:

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Originally Posted by abachler

Personally I woudl do this ---

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What's with the poor indentation?
Perhaps you need to read
http://cpwiki.sf.net/Indentation

Elysia, I did it like that but it wouldn't work. It didn't give me any errors, it just wouldn't abort when I entered 0.

The use of a loop in your code is not needed. I just demonstrated how the keyword break works.

I mean I used it in one of the first drafts of the program. I'll keep experimenting and see what I can come up with.

Thanks for the information you all have given :)

If abachler's code confused you a little, here it is a little simpler. It will take a bit longer to run though.

Code:

---

#include <iostream>
using namespace std;
 
// Prime number function - Tests number for remainder. If none is found, return false. Otherwise, return true.
 
bool PrimeFunction()
{
    int n;
 
    cin >> n;
    cin.ignore();
 
    for (int i = 2; i < n; i++)
    {
        if (n%i == 0)
        {
            cout << "Number is not prime";
            return false;
        }
    }
    cout << "Number is prime";
    return true;
}

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The code works by testing every factor of a number to see if it has no remainder, and if one such number evenly divides "n," n cannot be prime. You could replace "i < n" with "i * i < n", because you only need to test the factors up to the square root of n. If a factor of n is greater than the square root of n, then its corresponding factor is less than "n" so you would have found it anyways. IE, if n = 15, then you wouldn't need to test the factor "5" because the factor "3" would have been tested already.

Quote:

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Originally Posted by abachler

the original way will not accurately determine if a number is prime, it will only determine if a number is odd, not all odd numbers are prime ( 9 and 15 for example). The method I posted will absolutely determine if its prime or not. Although the method I gave you is probably the slowest 'fast' method there is, its also the simplest. There are far faster methods otu there, but the maths gets thick.

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I think he may have been referring to this line:

Code:

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    if (((n/temp)*temp) == n){

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It could be optimised and simplified to this:

Code:

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    if (n % temp == 0){

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Your while loop also involves a multiplication upon each loop, whereas sqrt(n) could be precalculated instead, saving that multiplication.

as I said, its probably the slowest 'fast' method out there, not really optimized at all other than to only test factrs less than the sqrt of the number in question. When I said there were faster methods, i wasnt refering to code level optimizations, btu rather algorithms that inherently prove or disprove primality faster. For example, not only will any composite number have a factor less than or equal to its square root, it will have a prime factor less than its square root. The proof is that any non-prime factor can itself be factored into smaler numbers, which will either be prime or can e further factored until some prime number is found. Any factor of a number A is also a factor of some number A * B.

A much faster method then is to only check the prime numbers less than or equal to sqrt(X). This is the method I used to generate the list of all 32 bit primes. The file doesnt in fact contain the number 2 although 2 is prime. Most algorithms exclude even numbers outright, so this isnt really an issue.

Actually that's about the slowest implementation of the slowest algorithm.
For algorithmic improvements you'd need to look into:
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
and
http://en.wikipedia.org/wiki/Sieve_of_Atkin
and more.

Not meaning to pick on you though. My main point is that it was written in an unnecessarily complicated way as recognised by the OP.