From: Sean Proctor <sproctor AT ccs DOT neu DOT edu>
Newsgroups: comp.os.msdos.djgpp
Subject: Re: prime numbers
Message-ID: <0br9ls8epo0j6qabj75jfg7f829p43ip1i@4ax.com>
References: <8j2kh1$v5e$1 AT nnrp1 DOT deja DOT com>
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Date: Sat, 24 Jun 2000 18:19:23 GMT
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To: djgpp AT delorie DOT com
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Reply-To: djgpp AT delorie DOT com

On Sat, 24 Jun 2000 15:36:01 GMT, kknd_bl AT hotmail DOT com wrote:

>how can I write a program that displays all the prime numbers (numbers
>that can only be divided by themselves and one) from 1 to N ??
>
>
>Sent via Deja.com http://www.deja.com/
>Before you buy.

I don't think here's where you ask people to write your programs for
you, but I'll do this one cause I'm bored.

first off, one isn't a prime number, so you don't need to include
that, and you know two is... you'll want something like this.

#include <stdio.h>
#include <stdlib.h>

typedef prime_list struct PRIME_LIST /*I always forget the order here,
that might be wrong*/
struct PRIME_LIST {
	prime_list *next
	int num;
};

int get_number(int argc, char *argv[]);

int div_by_list(int num, prime_list *list);

void add_num(int num, prime_list *list_end);

int main(int argc, char *argv[])
{
	int num;
	struct prime_list *list, *end;
	num = get_number(argc, argv); /*read number from arg list, if
it's not there, take input for it*/
	/*add 2 to the beginning of the list*/
	list = (prime_list *)malloc(sizeof(prime_list));
	list->num = 2;
	list->next = null;
	end = list;
	/*okay, now start stepping through all numbers*/
	for(i = 3; i <= num; i++)
		if(!div_by_list(i, list)) /*div i by each number in
list, return whether it was divisible by any*/
			add_num(i, end); /*if it wasn't add it to the
list*/
	print_list(list); /*self explanatory*/
	return 0;
}

int get_number(int argc, char *argv[])
{
	int num;
	char str[256];
	if(argc) {
		num = atoi(argv[0]);
		if(argc > 1) {
			printf("Invalid input string.\n");
			if(num){
				printf("Using the first number.\n"):
				return num;
			}
		}
		if(num)
			return num;
	}
	gets(str); /* I forget if this takes an argument or if it
returns a pointer, you'll have to figure that out */
	num = atoi(str);
	return num;
}

int div_by_list(int num, prime_list *list)
{
	prime_list *temp;
	temp = list;
	while(temp) {
		/* if there's no remainder, return true.*/
		if(num % temp->num == 0)
			return 1;
		temp = temp->next;
	}
	/*if it got to the end of the list, return false*/
	return 0;
}

void add_num(int num, prime_list *list_end)
{
	/*start at the end because lower numbers are more common and
so we don't have to step through the list*/
	list_end->next = (prime_list *)malloc(sizeof(prime_list));
	list_end = list_end->next;
	list_end->num = num;
	list_end->next = null;
}

if there's any problems, email me at sproctor AT ccs DOT neu DOT edu I haven't
tested it at all so it's likely there will be.  as far as I know that
should work though.

Sean