【MIT-OS6.S081作业1.3】Lab1-utilities primes

本文记录MIT-OS6.S081 Lab1 utilities 的primes函数的实现过程

1. 作业要求

primes (moderate)/(hard)

Write a concurrent version of prime sieve using pipes. This idea is due to Doug McIlroy, inventor of Unix pipes. The picture halfway down this page and the surrounding text explain how to do it. Your solution should be in the file user/primes.c.

Your goal is to use pipe and fork to set up the pipeline. The first process feeds the numbers 2 through 35 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 35.

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 35.
Once the first process reaches 35, 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.
• Your solution is correct if it implements a pipe-based sieve and produces 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
$

2. 实现过程

2.1 代码实现

作业提供了算法的链接：
Bell Labs and CSP Threads

先读取左边传来的所有数字，打印出第一个数字p，然后继续循环read左边给到的数字n，如果p不能被n整除，那么就把n继续发送给右边。

为了实现这个功能，我们这里需要使用fork和管道，fork有两个进程：主进程和子进程，我们分析一下它们分别是怎么做的：

• 第1个fork创建主进程$p_1$和子进程$n{p}_1$，从前面的ping-pong实验我们知道，主进程和子进程可以一个实现读，一个实现写，这是通过管道来实现的，于是我们创建一个管道$f{d}_1$（分别有$f{d}_1[0]和f{d}_1[1]$）。
• 主进程$p_1$将2~35写到第1个管道的写描述符$f{d}_1[1]$。
• 子进程$n{p}_1$从第1个管道的读描述符$f{d}_1[0]$中read第一个数字打印：2，为了将主进程发来的剩余数字继续发送到右边，必然不是写到$f{d}_1[1]$，这样和上面图片的数据方向就不一样了。我们需要在子进程创建新的管道$f{d}_2$，同时后面的逻辑还是类似的为主进程和子进程读写，我们依旧使用fork来创建得到主进程$p_2$和子进程$n{p}_2$（这里的$n{p}_1$和$p_2$是一样的），于是我们可以把主进程$p_1$发来的剩余数字在子进程的主进程$p_2$里写到第2个管道的写描述符$f{d}_2[1]$，在子进程$n{p}_2$来处理同样的读写，后面继续进行这样的读写操作…
• 我们可以使用递归【这样也就实现了提示说的You should create the processes in the pipeline only as they are needed.】。
• 递归停止条件：子进程从管道的读描述符读取第一个数字时返回0【read returns zero when the write-side of a pipe is closed.】，那么递归终止，递归过程在递归函数里建立新的管道，但是读取数字需要知道前一个管道，所以递归传参是前一个管道描述符的指针（指针传参）。

注意事项：

• write可以直接写入int类型的数字【It’s simplest to directly write 32-bit (4-byte) ints to the pipes, rather than using formatted ASCII I/O.】，传入int类型的指针即可。
• 主要关闭不需要的管道描述符号。

#include "kernel/types.h"
#include "user/user.h"void processPrime(int* p)
{close(p[1]);int num;if (0 == read(p[0], &num, sizeof(num))){close(p[0]);exit(0);}printf("prime %d\n", num);// create new pipeint nPipe[2];pipe(nPipe);int ret = fork();if (ret == 0){processPrime(nPipe);}else{close(nPipe[0]);int nNum;while (read(p[0], &nNum, sizeof(nNum))){if (nNum % num != 0)write(nPipe[1], &nNum, sizeof(nNum));}close(p[0]);close(nPipe[1]);wait(0);}exit(0);
}
int main(int argc, char* argv[])
{int p[2];pipe(p);int ret = fork();if (ret == 0){processPrime(p);}    else{close(p[0]);for (int i = 2; i <= 35; ++i)write(p[1], &i, sizeof(i));close(p[1]);wait(0);}exit(0);
}

我们在Makefile里加入对这个函数的编译：

UPROGS=\$U/_cat\$U/_echo\$U/_forktest\$U/_grep\$U/_init\$U/_kill\$U/_ln\$U/_ls\$U/_mkdir\$U/_rm\$U/_sh\$U/_stressfs\$U/_usertests\$U/_grind\$U/_wc\$U/_zombie\$U/_sleep\ $U/_pingpong\ $U/_primes\ 

重新编译通过，测试primes，如题所述正确打印：

确实过了一会然后可以继续输入命令，然后我们再使用作业所说的测试命令：

./grade-lab-util primes

完活！

递归来写还是比较清晰的。