Thread: Here is my prime code! why wont it work?

The prime number is passed into this function. This works fine for 7 but when I enter 407 it tells me it is prime when it is not.

Here goes

char test_prime(int prime)

{

int r; /*remainder*/
int i;

for (i =2; i < prime;i++)
if (prime % i == 0)
{
return('n'); /*not prime*/
}
else
{
return('p');
}

Thanks guys
Joe

Your code does not work for prime=9, it returns 'p'.
Your "if ... else" is not good.

For i=2
9 % i = 1 != 0
so your prog goes to "else" and returns 'p' !!!
Just take off the "else" keyword ...

In fact your prog only looks for odd numbers returning 'p'.

some hints : your loop can be shorter
for (i =2; i < prime/2;i++)

for (i =2; i < sqrt(prime)+1;i++)

excepted i=2, you can consider only odd i.

cprogramnewbie
Your loop is no loop. As GertFaller stated, when the program goes into the for-loop, it immediately returns.

Much easier would be to extend your condition. The example below shows how such is done, in case you didn't know such is possible.

Further you should use return only at the end of your function. In general this makes a program much easier to read and to understand. I always introduce a return value which value is set by the code in the function and returned at the end.

Code:

is_prime = TRUE
for (i = 2; i < prime && is_prime == TRUE; i++)
{
    if (prime % i == 0)
        is_prime = FALSE;
}

return (is_prime == FALSE) : 'n' ? 'p';

Gugge

Also to you, don't use so much returns in your function. You also have logic error in your code. In your while loop, which is not really a while loop, an if-else construction is used which results in an immediate return. So the last return in the function will never be reached. To avoid problems like this, it is better to keep a return value.

Further, assume

prime = 9
i = 2

then

9 % 2 == 1

So your function says: that it is a prime?