使用 fork() 的 C 程序的行为

Question

Given the following code, I have to check its behavior, that is, how much processes are produced.给定以下代码，我必须检查它的行为，即产生了多少进程。

#include <unistd.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char*argv[])
{
    int pid, i , j;
    pid_t t;
    for (i = 0; i < 2; i++) {
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
            t = fork();
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }

    printf("I am the process %d and my father is the process %d\n", getpid(), getppid());

    while(wait(NULL)>0) {}

    return 0;
}

After compiling and executing, the output is as follows:编译执行后output如下：

I am the process 3749 and my father is the process 2254

I am the process 3750 and my father is the process 3749

I am the process 3751 and my father is the process 3749

I am the process 3752 and my father is the process 3749

I am the process 3754 and my father is the process 3749

I am the process 3753 and my father is the process 3750

I am the process 3755 and my father is the process 3750

I am the process 3756 and my father is the process 3752

I am the process 3757 and my father is the process 3752

I am the process 3758 and my father is the process 3756

I am the process 3759 and my father is the process 3756

This means the corresponding tree process looks like this:这意味着相应的树过程如下所示：

                      2254
                       |
                   -- 3749 ----
                  /   |   \    \
                3750 3751 3752 3754
               /   \      /  \
            3753  3755  3756 3757
                       /   \
                     3758  3759

In total, 11 processes are produced, if we count the initial process.如果我们计算初始进程，总共会产生 11 个进程。

I expected much more processes to be generated than just 11;我预计会生成更多的进程，而不仅仅是 11 个； anyway, I am able to understand the spawning of all processes but the most distant leaves (3758 and 3759).无论如何，我能够理解除了最远的叶子（3758 和 3759）之外的所有进程的产卵。

In the program there are two for loops, being one nested into the other.在程序中有两个 for 循环，一个嵌套在另一个中。 As two break statements are inside each loop, I guess that a loop or the two should finish at some point before their natural end.由于每个循环中有两个 break 语句，我猜一个或两个循环应该在其自然结束之前的某个时间点完成。

Could somebody confirm this code's behavior?有人可以确认此代码的行为吗？

Thank you in advance先感谢您

Answer 1

These kinds of programs are more to illustrate that you understand how forking works, I hope you never write a forking program within multiple nested loops complete with breaks.这些类型的程序更多是为了说明您了解分叉是如何工作的，我希望您永远不要在多个嵌套循环中编写分叉程序，其中包含中断。

They way you track this is by simulating the machine on paper, but each time you simulate the fork, you copy the simulation, setting the return code of fork() for one copy to zero, and the return code of the other copy to the other process's pid.他们跟踪这一点的方法是通过在纸上模拟机器，但是每次模拟分叉时，您都会复制模拟，将fork()的返回代码设置为一个副本为零，并将另一个副本的返回代码设置为其他进程的pid。

Let me reformat your program to make it easier让我重新格式化您的程序以使其更容易

    int pid;
    int i;
    int j;
    pid_t t;

will hold our "current values" for a process.将保存我们的“当前值”的过程。 We will also indicate the instruction pointer location with "IP ==>"我们还将用“IP ==>”指示指令指针位置

    int pid;
    int i;
    int j;
    pid_t t;
    IP ===>
    for (i = 0; i < 2; i++) {
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
            t = fork();
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }

Next step (for PID 1)下一步（对于 PID 1）

    === PROCESS 1 ====
    int pid;
    int i; (set to 0)
    int j;
    pid_t t;

    for (i = 0; i < 2; i++) {
    IP ===>
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
            t = fork();
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }

next step (for PID 1)下一步（对于 PID 1）

    === PROCESS 1 ====
    int pid (set to 1  // fake pid, but easy to count)
    int i; (set to 0)
    int j;
    pid_t t;

    for (i = 0; i < 2; i++) {
        pid = getpid();
    IP ===>
        for (j = 0 ; j < i+2; j++){
            t = fork();
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }

next step (for process 1)下一步（用于流程 1）

    === PROCESS 1 ====
    int pid (set to 1)
    int i; (set to 0)
    int j; (set to 0)
    pid_t t;

    for (i = 0; i < 2; i++) {
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
    IP ===>
            t = fork();
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }

next step (for PID 1)下一步（对于 PID 1）

    === PROCESS 1 ====
    int pid (set to 1)
    int i; (set to 0)
    int j; (set to 0)
    pid_t t; (set to 2, parent process)

    for (i = 0; i < 2; i++) {
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
            t = fork();
    IP ===>
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }
    === PROCESS 2 ===
    int pid (set to 1)
    int i; (set to 0)
    int j; (set to 0)
    pid_t t; (set to 0, child process)

    for (i = 0; i < 2; i++) {
        pid = getpid();
        for (j = 0 ; j < i+2; j++){
            t = fork();
    IP ===>
            if (t != 0) {
                t = fork();
                break;
            }
        }
        if (pid != getpid())
            break;
    }