Primes

Your goal is to use pipe and fork to set up the pipeline. The first process feeds the numbers 2 through 280 into the pipeline. For each prime number, you will arrange to create one process that reads from its left neighbor over a pipe and writes to its right neighbor over another pipe. Since xv6 has limited number of file descriptors and processes, the first process can stop at 280.

Some hints:

• Be careful to close file descriptors that a process doesn't need, because otherwise your program will run xv6 out of resources before the first process reaches 280.

• Once the first process reaches 280, it should wait until the entire pipeline terminates, including all children, grandchildren, &c. Thus the main primes process should only exit after all the output has been printed, and after all the other primes processes have exited.

• Hint: read returns zero when the write-side of a pipe is closed.

• It's simplest to directly write 32-bit (4-byte) ints to the pipes, rather than using formatted ASCII I/O.

• You should create the processes in the pipeline only as they are needed.

• Add the program to UPROGS in Makefile.

• If you get an infinite recursion error from the compiler for the function primes, you may have to declare void primes(int) __attribute__((noreturn)); to indicate that primes doesn't return.

Your solution should implement a pipe-based sieve and produce the following output:

    $ make qemu
    ...
    init: starting sh
    $ primes
    prime 2
    prime 3
    prime 5
    prime 7
    prime 11
    prime 13
    prime 17
    prime 19
    prime 23
    prime 29
    prime 31
    ...
    $

#include "kernel/types.h"
#include "kernel/stat.h"
#include "user/user.h"
#define RD 0
#define WR 1

int main(int argc, char *argv[])
{
	int pp[2];	//pipe_port
	int prime;
	int data;
	int pid=0;

	// initialize
	pipe(pp);
	for (int i=2; i<=35; i++)
		write(pp[WR], &i, sizeof(int));
	close(0);
	dup(pp[RD]);
	close(pp[RD]);
	close(pp[WR]);
	prime=2;
	printf("prime 2\n");

	// every sub-process execute the same logic
	while (read(0, &data, sizeof(int))) {
		if (data % prime != 0) {
			if (!pid) {	//is a child
				pipe(pp);
				pid=fork();
				if (pid) {	//parent
					close(pp[RD]);
				} else {	//child
					close(0);
					dup(pp[RD]);
					close(pp[RD]);
					close(pp[WR]);
					prime=data;
					printf("prime %d\n", data);
				}
			}
			write(pp[WR], &data, sizeof(int));
		}
	}
	close(pp[WR]);
	wait(0);
	exit(0);
}