Programming Forums > Application Development > C

Cloudbreak · Nov 1st, 2004, 9:28 AM

Hello, I'm quite new to the C language and I wanted to try and make a program that'll input a number from the user and tell him whether it is prime or not.
This is the program I tried to build:

Quote:

#include <stdio.h>
#include <conio.h>
#include <math.h>
int main()
{
int i, number, square;

printf("Enter a potential prime:\n");
scanf("%d", &number);

square = sqrt(number);

for (i=2; i < square; i++)
{
if(number%i==0)
printf("Number is prime.\n");
else
printf("Number is not prime because it is divisible by %d", i);
}

getch();
return 0;
}*

No matter what number I input it tells me it is divisible by 2. Please help me if you know how to fix it...

Thanks ahead.
Cloudbreak.

Pizentios · Nov 1st, 2004, 10:57 AM

First off, last time i checked a prime number was defined like this:

A prime number is a positive integer that has exactly two positive integer factors, 1 and itself. For example, if we list the factors of 28, we have 1, 2, 4, 7, 14, and 28. That's six factors. If we list the factors of 29, we only have 1 and 29. That's 2. So we say that 29 is a prime number, but 28 isn't.Another way of saying this is that a prime number is a whole number that is not the product of two smaller numbers. Note that the definition of a prime number doesn't allow 1 to be a prime number: 1 only has one factor, namely 1. Prime numbers have exactly two factors, not "at most two" or anything like that. When a number has more than two factors it is called a composite number.

(Toggle Plain Text)

#include <iostream.h>

int main()
{
	int num;
	cout << "\nPlease enter a number to check: ";
	cin >> num;
	
	if ((num%2) == 0)
	{
 cout << "\nNumber is not prime.\n";
	}
	else
	{
 cout << "\nNumber is prime.\n";
	}
	return 0;
}

This is how i'd do it. works for every prime except 2. if you want a list of prime numbers to test with there is a great list here. It's the first 1000 prime numbers. I think you were getting mixed up witht eh square root part, you really don't need it unless your trying to output a whole bunch of prime numbers....not checking for just one. I could be wrong though. Can somebody back me up on this?

Ooble · Nov 1st, 2004, 11:36 AM

That above example would state that numbers such as 9, 15 and 27 were prime - they're not. If you want to check primes, try the following:

(Toggle Plain Text)

int prime = 1;

for (int i = 0; i < sqrt(number); i++)
{
  if ((number / i) == (int)(number / i))
    /* i is a factor - the number divided by i is the same when converted to an integer. */
    prime = 0;
}

if (prime == 0)
  printf("%i is not a prime number.", number);
else
  printf("%i is a prime number.", number);

Not tested, but it should work.

Mjordan2nd · Nov 1st, 2004, 4:12 PM

Here's how I did it:

(Toggle Plain Text)

#include <stdio.h>
#define TRUE 1
#define FALSE 0

int main(void)
{
	printf("enter possible prime number");
	int number;
	scanf("%d", &number);
	
	int prime = TRUE;
	int i;
	for(i = 2; i<number; i++)
	{
 if(number%i==0)
 {
 	prime = FALSE;
 }
	}
	if(number == 2)
	{
 prime = TRUE;
	}
	
	if(prime)
	{
 printf("%d is a prime number", number);
	}
	else
	{
 printf("%d isn't a prime number", number);
	}
	return 0;
}

Pizentios · Nov 2nd, 2004, 9:56 AM

Quote:

That above example would state that numbers such as 9, 15 and 27 were prime - they're not.

Correct, i didn't do much testing on that code. Heh, now i look like an idiot.

Ooble · Nov 2nd, 2004, 11:48 AM

Mine's slightly buggered - it should be:

(Toggle Plain Text)

  if ((number / i == (int)(number / i)) && i != 1 && i != number)
    prime = 0;
    /* i is a factor that is not 1 or the number itself - the number divided by i is the same when converted to an integer. */
  if (number == 1)
    prime = 0;
    /* 1 is not a prime number */

If prime is still 1 after all that, the number's a prime.

Nov 1st, 2004, 10:57 AM	#2
Pizentios Programming Guru Join Date: May 2004 Location: Brandon, Manitoba, Canada Posts: 2,023 Rep Power: 7	First off, last time i checked a prime number was defined like this: A prime number is a positive integer that has exactly two positive integer factors, 1 and itself. For example, if we list the factors of 28, we have 1, 2, 4, 7, 14, and 28. That's six factors. If we list the factors of 29, we only have 1 and 29. That's 2. So we say that 29 is a prime number, but 28 isn't.Another way of saying this is that a prime number is a whole number that is not the product of two smaller numbers. Note that the definition of a prime number doesn't allow 1 to be a prime number: 1 only has one factor, namely 1. Prime numbers have exactly two factors, not "at most two" or anything like that. When a number has more than two factors it is called a composite number. (Toggle Plain Text) #include <iostream.h> int main() { int num; cout << "\nPlease enter a number to check: "; cin >> num; if ((num%2) == 0) { cout << "\nNumber is not prime.\n"; } else { cout << "\nNumber is prime.\n"; } return 0; } #include <iostream.h> int main() { int num; cout << "\nPlease enter a number to check: "; cin >> num; if ((num%2) == 0) { cout << "\nNumber is not prime.\n"; } else { cout << "\nNumber is prime.\n"; } return 0; } This is how i'd do it. works for every prime except 2. if you want a list of prime numbers to test with there is a great list here. It's the first 1000 prime numbers. I think you were getting mixed up witht eh square root part, you really don't need it unless your trying to output a whole bunch of prime numbers....not checking for just one. I could be wrong though. Can somebody back me up on this? __________________ Profanity is the one language that all programmers understand. Check out my Blog <---updated Nov 30 2007!

Nov 1st, 2004, 11:36 AM	#3
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	That above example would state that numbers such as 9, 15 and 27 were prime - they're not. If you want to check primes, try the following: (Toggle Plain Text) int prime = 1; for (int i = 0; i < sqrt(number); i++) { if ((number / i) == (int)(number / i)) /* i is a factor - the number divided by i is the same when converted to an integer. / prime = 0; } if (prime == 0) printf("%i is not a prime number.", number); else printf("%i is a prime number.", number); int prime = 1; for (int i = 0; i < sqrt(number); i++) { if ((number / i) == (int)(number / i)) / i is a factor - the number divided by i is the same when converted to an integer. */ prime = 0; } if (prime == 0) printf("%i is not a prime number.", number); else printf("%i is a prime number.", number); Not tested, but it should work. __________________ Me :: You :: Them

Nov 1st, 2004, 4:12 PM	#4
Mjordan2nd The Supreme Ruler Join Date: May 2004 Location: Houston Posts: 1,476 Rep Power: 6	Here's how I did it: (Toggle Plain Text) #include <stdio.h> #define TRUE 1 #define FALSE 0 int main(void) { printf("enter possible prime number"); int number; scanf("%d", &number); int prime = TRUE; int i; for(i = 2; i<number; i++) { if(number%i==0) { prime = FALSE; } } if(number == 2) { prime = TRUE; } if(prime) { printf("%d is a prime number", number); } else { printf("%d isn't a prime number", number); } return 0; } #include <stdio.h> #define TRUE 1 #define FALSE 0 int main(void) { printf("enter possible prime number"); int number; scanf("%d", &number); int prime = TRUE; int i; for(i = 2; i<number; i++) { if(number%i==0) { prime = FALSE; } } if(number == 2) { prime = TRUE; } if(prime) { printf("%d is a prime number", number); } else { printf("%d isn't a prime number", number); } return 0; } __________________ "Every gun that is made, every warship launched, every rocket signifies, in the final sense, a theft from those who hunger and are not fed, from those who are cold and are not clothed. The world in arms is not spending money alone. It is spending the sweat of its laborers, the genius of its scientists, the hopes of its children." - Dwight D. Eisenhower

Nov 2nd, 2004, 11:48 AM	#6
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	Mine's slightly buggered - it should be: (Toggle Plain Text) if ((number / i == (int)(number / i)) && i != 1 && i != number) prime = 0; /* i is a factor that is not 1 or the number itself - the number divided by i is the same when converted to an integer. / if (number == 1) prime = 0; / 1 is not a prime number / if ((number / i == (int)(number / i)) && i != 1 && i != number) prime = 0; / i is a factor that is not 1 or the number itself - the number divided by i is the same when converted to an integer. / if (number == 1) prime = 0; / 1 is not a prime number */ If prime is still 1 after all that, the number's a prime. __________________ Me :: You :: Them

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Programming Forums > Application Development > C

Prime Numbers In C