父局部变量充当三个孩子之间的共享变量

Question

在下面的程序中，父进程的局部变量如何成为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：

进程 0 运行

进程 1 运行

进程 2 运行

但根据我的说法，最后两个进程应该挂起，因为 turn 的值不应该改变它们。

如果我放了一个 exit(0);在转++之后，我立即返回到我的shell提示符，只有一行输出，即： 进程 0 运行

但其他两个进程仍然在后台运行，其他两个进程没有任何输出。

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，您需要记住，您将继续运行 for 循环的所有进程，而不仅仅是主进程。在那之后，弄清楚发生了什么并不难。您总共生成了 6 个进程，其中三个将具有正确的状态 i 和 turn。

Answer 2

像这样重构你的代码：

#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;
}

这可能会让您感到震惊，但上面的 Art 重构非常清楚地表明您的代码将产生 6 个进程。在其中添加更多 printfs 应该可以让您了解真正发生的事情。

Answer 3

在 fork() 之前声明的变量是为子进程共享的。您可以通过打印 &turn in loop 来检查它。会一样的。