While loop help!

8100 Power · #1 09-Feb-2009, 19:50

Ok, I'm writing a program to determine if the number entered is prime. So, here is my main script:

CPP / C++ / C Code:

cout << "Please enter digits to determine if prime or not, use 0 to quit"<< endl;
    cin >> input;
    
    while( input != 0 && input > 0){
       
       cin >> input;
       
       while(input % 2){   // <---This is my problem?? reference below!
          
          if ( input == 1)
              cout << "The Number is Prime" << endl;
          else
              cout << "The Number is not prime" << endl;
       }
             
    }

You can see where I pointed out my problem ( I think). Can I input a arthmatic operator here to process every # inputted to determine the if condition?

Basically, every prime number modulous 2 puts out a 1, so I can determine if the # is prime or not by the if condition.

Thanks for any help.

ocicat · #2 09-Feb-2009, 22:21

Quote:

Originally Posted by 8100 Power

Basically, every prime number modulous 2 puts out a 1...

This is not entirely correct. While it is true that even numbers are not prime (depending upon whether 2 is considered prime or not...), for all values x where x % 2 == 1 really is just another definition of the odd number sequence.

Consider the value 9. 9 % 2 == 1, but 9 is not prime.

8100 Power · #3 09-Feb-2009, 23:15

Ok.

2 questions:

1. What would be the most simplest and correct way to determine if the number is prime?

2. Is my code correct?

8100 Power · #4 09-Feb-2009, 23:41

As I've wrote my code a little different, I've tested all numbers and everything is coming out as expected. Now, I just need to integrate my while loops. What is wrong (programming wise) with my code?

Howard_L · #5 10-Feb-2009, 00:11

What code?
You know, finding prime has been covered here before.
It might help you to search the forum for previous threads.

8100 Power · #6 10-Feb-2009, 09:21

Quote:

Originally Posted by Howard_L

What code?
You know, finding prime has been covered here before.
It might help you to search the forum for previous threads.

I'm now asking for help about my while loop, not process of determining a prime number.

The code in the first reply if you would of read it.

ocicat · #7 10-Feb-2009, 10:31

Quote:

Originally Posted by 8100 Power

What would be the most simplest and correct way to determine if the number is prime?

For any number n, n is prime if all prime numbers less than n are found to not be factors of n.

Quote:

Is my code correct?

No. Your code asks for user input twice followed by checking to see if the last value received is odd & then outputs strings based on whether the value is 1 or not. This does not follow the definition of prime numbers.

8100 Power · #8 10-Feb-2009, 20:55

Quote:

Originally Posted by ocicat

For any number n, n is prime if all prime numbers less than n are found to not be factors of n.

No. Your code asks for user input twice followed by checking to see if the last value received is odd & then outputs strings based on whether the value is 1 or not. This does not follow the definition of prime numbers.

I'm still tying to register that sentence in my head. Could you elaborate?

As for my while structure. When the user enters the values, shouldn't that be in the outer loop?

Howard_L · #9 10-Feb-2009, 21:57

Quote:

The code in the first reply if you would of read it.

Sorry but you said:

Quote:

As I've wrote my code a little different

Does the new code still get user input twice?
Does the new code still rely on this to determine a prime number? :

CPP / C++ / C Code:

while(input % 2)

Hopefully not. So you see, it would be helpful if you would post your newest code. That's all I was trying to say...

ocicat · #10 10-Feb-2009, 22:19

Quote:

Originally Posted by 8100 Power

I'm still tying to register that sentence in my head.

A prime number is only divisible by 1 & itself.

So look at a number of examples:

The reason 9 is not prime is because it can be factored to 3 * 3.
All even numbers are divisible by 2, & given that 2 is prime, this means that all even numbers cannot be prime (Again, primes are divisible only by themselves & 1...).
11 is only divisible by itself & 1, so 11 is a prime number.
21 is not prime because it is divisible by 3 & 7.

So you should beginning to understand that odd numbers aren't necessarily prime. Some are; others aren't.

Quote:

As for my while structure. When the user enters the values, shouldn't that be in the outer loop?

That is one alternative. You could also exit when the input is not numeric or negative, however, this is not overly pertinent to determining whether a value has a prime factorization or not.

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