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Old October 9th, 2006, 11:09 PM
b.chuch b.chuch is offline
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An Array of Natural Numbers

Hi, I'm new to this forum, and would really appreciate some assistance. I'm trying to create a program that tests if a number is prime or not. In order to do that, I'm going to need an array containing all natural numbers up to some obscenly large number. I then need to find a way to test the number against all elements of the array, to see if after the decimal place, there is a 0. How would this be done? Thanks in advance

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Old October 10th, 2006, 05:23 AM
costas costas is offline
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Well, I think you don't have to get it to so much trouble as I presume that the "math.h" library must have some function or something to test if a number is prime or not. Search in Google or something! Hope I helped!!!

Costas

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Old October 10th, 2006, 05:41 AM
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Hmmm. Not a very efficient method. There are many algorithms that are much more effective, but the simplest (though not necessarily the fastest is something like this:

C Code:
Original - C Code
  1. #include <stdio.h>
  2.  
  3. int is_prime(int);
  4.  
  5. int main()
  6. {
  7.         int i;
  8.         for(i = 1; i < 20; i++)
  9.         {
  10.                 if(is_prime(i) == 0)
  11.                         printf("%d is a prime\n", i);
  12.                 else
  13.                         printf("%d is not a prime\n", i);
  14.         }
  15.  
  16.         return 0;
  17.  
  18. }
  19.  
  20. int is_prime(int number)
  21. {
  22.         int i;
  23.         for(i = 2; i <= (number / 2); i++)
  24.         {
  25.                 if (((number / i) * i) == number)
  26.                         return i;
  27.         }
  28.         return 0;
  29. }
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This is my code. Is it not nifty?

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