Question

如何在下面的程序中，父进程的局部变量充当三个子进程之间的共享变量。

int main()
{
    int turn = 0;
    int i;

    for (i = 0; i < 3; i++)
    {
        if (fork() == 0)
        {
            int me = i;
            while (turn != me)
                /*do nothing*/ ;

            // my turn
            printf("Process %d ran\n", me);
            turn++;
        }
    }

    return 0;
}

输出：

流程0运行

流程1运行

流程2运行

但据我所知，最后两个进程应该挂起，因为turn的值永远不会改变它们。

如果我放一个出口（0）;在转换++之后，我立即返回到我的shell提示符，只有一行输出，即：流程0运行

但是其他两个进程仍然在后台运行，而其他两个进程都没有输出。

Answer 1

发生了什么：

main process is 0.

process 0 calls fork. i = 0;
process 0 returns from fork = 1;
process 0 calls fork. i = 1;
process 0 returns from fork = 2;
process 0 calls fork. i = 2;
process 0 returns from fork = 3;
process 0 exists;

process 1 returns from fork. i = 0;
process 1 hits the while loop turn = 0, i = 0, me = 0, while loop exists;
process 1 calls printf.
process 1 increments turn, turn = 1;
process 1 goes back to the for loop;
process 1 calls fork, i = 1;
process 1 returns from fork = 4;
process 1 calls fork, i = 2;
process 1 returns from fork = 5;
process 1 exists;

process 2 returns from fork. i = 1;
process 2 hits the while loop turn = 0, i = 1, me = 0, while loop spins forever;

process 3 returns from fork, i = 2;
process 3 hits the while loop turn = 0, i = 2, me = 0, while loop spins forever;

process 4 (spawned from process 1) returns from fork, i = 1, turn = 1 (inherited from process 1).
process 4 hits the while loop turn = 1, i = 1, me = 1, while loop exits;
process 4 calls printf;
process 4 increments turn, turn = 2;
process 4 goes back to the for loop;
process 4 calls fork, i = 2, turn = 2;
process 4 returns from fork = 6;
process 4 exits;

process 5 (spawned from process 1) return from fork, i = 2, turn = 1 (inherited from process 1)
process 5 hits the while loop turn = 1, i = 2, me = 2, while loop spins forever;

process 6 (spawned from process 4) returns from fork i = 2, turn = 2 (inherited from process 4)
process 6 hits the while loop turn = 2, i = 2, me = 2, while loop exists;
process 6 calls printf;
process 6 drops out of the for loop
process 6 exits;

基本上，你需要记住你将继续运行for循环的所有进程，而不仅仅是主进程。在那之后，要弄清楚发生了什么并不是很难。您总共产生了6个进程，其中3个进程的状态为i和turn。

Answer 2

重构您的代码：

#include <unistd.h>
#include <stdio.h>

int main()
{
    int turn = 0;
    int i;

    for (i = 0; i < 3; i++)
    {
        printf("forking i = %d, turn = %d\n", i, turn);

        if (fork() == 0)
        {
            int me = i;
            while (turn != me)
                /*do nothing*/ ;

            // my turn
            printf("Process %d ran\n", me);
            turn++;
        }
    }

    return 0;
}

这可能会给你带来震撼，但正如上面的Art重构显示你的代码会产生6个进程。在那里放置更多的printfs应该让你了解真正发生的事情。

Answer 3

在子进程共享fork（）之前声明的变量。您可以通过打印和循环来检查它。它会是一样的。

父本地变量充当三个孩子之间的共享变量

3 个答案: