primes.c

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

#define LOW_PRIME_COUNT 6542

int main()
{
	puts("2\n3");

	unsigned primes[LOW_PRIME_COUNT];

	memset(primes, 255, sizeof(primes));

	primes[0] = 2;
	primes[1] = 3;

	int index = 2;

	unsigned a = 2, a1 = 3, a2 = 9, current = 5; //count = 1;

	for (;;) {
		int i;

		for (i = 2; primes[i] < a1; i++) {
			if (!(current % primes[i])) {
				goto notaprime;
			}
		}

		if (index < LOW_PRIME_COUNT) {
			primes[index] = current;
			++index;
		}

		printf("%u\n", current);

notaprime:

		if (current == 0xFFFFFFFD) {
			break;
		}

		current += a;
		a ^= 6;

		if (current >= a2) {
			if (current == a2) {
				current += a;
				a ^= 6;
			}
			a2 += a1;
			++a1;
			a2 += a1;
		}
	}

	exit(EXIT_SUCCESS);
}

primes2.c

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

int main(int argc, char *argv[])
{
	unsigned *sieve = calloc(1 << 28, 1);

	if (!sieve) {
		err(EXIT_FAILURE, "Failed to allocate memory for a sieve");
	}

	puts("2");

	unsigned a = 3, b = 0;

	do {
		if (!(sieve[b >> 5] & (1 << (b & 0x1F)))) {
			printf("%d\n", a);
			unsigned c = a + b;

			do {
				sieve[c >> 5] |= 1 << (c & 0x1F);
				c += a;
			}	while (c < 0x80000000);
		}
		++b;
		a += 2;
	} while (a < 0x10000);

	do {
		if (!(sieve[b >> 5] & (1 << (b & 0x1F)))) {
			printf("%u\n", a);
		}
		++b;
		a += 2;
	} while (a < 0xFFFFFFFF);

	free(sieve);

	exit(EXIT_SUCCESS);
}