August 24th, 2013, 01:22 AM

For loop for test prime numbers
Hi, In the following code, a test for determining the primality of a number is made:
Code:
#include <stdio.h>
#include <math.h>
int main()
{
unsigned int n, FLAG_PRIME;
unsigned int n_square;
unsigned int x;
FLAG_PRIME = 1;
printf("\nUntil which limit?: ");
scanf("%d", &n);
n_square = (unsigned int)sqrt((double)n);
for(x = 2; x == n_square; ++x)
{
if (n % x == 0)
FLAG_PRIME = 0;
}
if (FLAG_PRIME == 1)
printf("%d prime\n", n);
else
printf("%d not prime\n", n);
return 0;
}
However, in the line
Code:
for(x = 2; x == n_square; ++x)
I don't know the value of x for numbers in which the square root of n (n_square) is shorter than x ie sqrt(2) = 1.414. It's a Little puzzling since it works but I would like to know the behavior of for loops in this particular cases.
Thanks in advanced
August 24th, 2013, 02:41 AM

If you want to know how it works (or more accurately, doesn't work), then do this.
Code:
for(x = 2; x == n_square; ++x)
{
printf("Checking if %d is a divisor of %d\n", x, n );
if (n % x == 0)
FLAG_PRIME = 0;
}
FYI, the for loop only runs while the test expression returns true.
Comments on this post
August 25th, 2013, 01:01 AM

Thanks (SOLVED)
Originally Posted by salem
If you want to know how it works (or more accurately, doesn't work), then do this.
Code:
for(x = 2; x == n_square; ++x)
{
printf("Checking if %d is a divisor of %d\n", x, n );
if (n % x == 0)
FLAG_PRIME = 0;
}
FYI, the for loop only runs while the test expression returns true.
Since for doesn't run in those cases and Flag 1 is the default, it appeared that worked. Your reply was very helpful and is greatly appreciated