我正在尝试在 C 中制作 fork() 进程调用图

Question

这是我现在的位置

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>

main(){   
    int n;
    int i;
    int x = 1;

    printf("\nenter number of forks:\n");
    scanf ("%d",&n);
    printf("\nnow forking %d times....\n\n", n);

    for (i = 1;i <= n; i++){
        int pid = fork();
        if (pid < 0){ 
            return -1;
        }

        if (pid != 0){
            printf("\nI am the parent (
                ppid = %d pid = %d)\n",getppid(),getpid()
            );          
            printf(" x = %d\n",x);

            x++;
            wait();
        }else{
            printf("\nI am the child (
                ppid = %d, pid = %d)\n x = 
                %d\n-------------------------------\n",
                getppid(),getpid(),x
            );
        }
    }
}

这是我传入 4 时的输出：

enter number of forks:
4

now forking 4 times....


I am the parent (ppid = 26178 pid = 39785)
 x = 1

I am the child (ppid = 39785, pid = 39786)
 x = 1
-------------------------------

I am the parent (ppid = 39785 pid = 39786)
 x = 1

I am the child (ppid = 39786, pid = 39787)
 x = 1
-------------------------------

I am the parent (ppid = 39786 pid = 39787)
 x = 1

I am the child (ppid = 39787, pid = 39788)
 x = 1
-------------------------------

I am the parent (ppid = 39787 pid = 39788)
 x = 1

I am the child (ppid = 39788, pid = 39789)
 x = 1
-------------------------------

I am the parent (ppid = 39786 pid = 39787)
 x = 2

I am the child (ppid = 39787, pid = 39790)
 x = 2
-------------------------------

I am the parent (ppid = 39785 pid = 39786)
 x = 2

I am the child (ppid = 39786, pid = 39791)
 x = 2
-------------------------------

I am the parent (ppid = 39786 pid = 39791)
 x = 2

I am the child (ppid = 39791, pid = 39792)
 x = 2
-------------------------------

I am the parent (ppid = 39785 pid = 39786)
 x = 3

I am the child (ppid = 39786, pid = 39793)
 x = 3
-------------------------------

I am the parent (ppid = 26178 pid = 39785)
 x = 2

I am the child (ppid = 39785, pid = 39794)
 x = 2
-------------------------------

I am the parent (ppid = 39785 pid = 39794)
 x = 2

I am the child (ppid = 39794, pid = 39795)
 x = 2
-------------------------------

I am the parent (ppid = 39794 pid = 39795)
 x = 2

I am the child (ppid = 39795, pid = 39796)
 x = 2
-------------------------------

I am the parent (ppid = 39785 pid = 39794)
 x = 3

I am the child (ppid = 39794, pid = 39797)
 x = 3
-------------------------------

I am the parent (ppid = 26178 pid = 39785)
 x = 3

I am the child (ppid = 39785, pid = 39798)
 x = 3
-------------------------------

I am the parent (ppid = 39785 pid = 39798)
 x = 3

I am the child (ppid = 39798, pid = 39799)
 x = 3
-------------------------------

I am the parent (ppid = 26178 pid = 39785)
 x = 4

I am the child (ppid = 39785, pid = 39800)
 x = 4
-------------------------------

我注意到的第一件事是，对于代码的每个实例运行，“子”的 PPID 是“父”的 PID，这很好。

但是当我手绘图的时候： diagram of my output (I'm new so I can't post photos)

为了作业的缘故，应该是一棵平衡树，这样层次的概念才有意义。我希望将每个进程打印为图表中的节点，例如我使用 graphviz 绘制的图表，并且每个节点都应包含其 PID 和级别。

这是极客对极客的一个例子，展示了我所说的级别的意思：


       L1       // There will be 1 child process 
    /     \     // created by line 1.
  L2      L2    // There will be 2 child processes
 /  \    /  \   //  created by line 2
L3  L3  L3  L3  // There will be 4 child processes 
                // created by line 3

我想我编码过度了。我应该如何更改我的代码以在循环中分叉它并获得树状结构？我正在尝试使用变量 x 来表示级别，但我不确定它是否有效，因为输出的结构对我来说非常意外。

Answer 1

你的直觉错了。在循环的每次迭代中，所有您的进程将分叉并创建一个新进程 - 不仅是在上一次迭代中创建的进程。

因此，经过 4 次迭代，您会期望初始过程以四个 children:

结束

一有3children（图中55）
一有2children（图中63）
一个有 1 个 child（图中为 67）；和
无child仁（图中69）

这正是您最终得到的结果。如果您对这些 children 中的每一个都遵循相同的逻辑，并适当减少剩余循环迭代的数量，您将完全重现您拥有的图表。

This question 并非完全相同，但涵盖了大部分相同的领域。

当然，您的图表是一棵树 - 它不是完全平衡的二叉树。要获得这样的结果，您必须在每次循环迭代期间按照以下方式执行更多操作：

分叉进程两次
在每个 children 中，立即继续下一个循环迭代（这样在下一个迭代之前它们都不会再次分叉）
在 parent 中，分叉后完全脱离循环，这样它就不会再次分叉，永远

例如：

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(void) {
    int level;

    for ( level = 1; level <= 3; level++ ) {
        pid_t pids[2];

        pids[0] = fork();
        if ( pids[0] == -1 ) {
            perror("fork failed");
            exit(EXIT_FAILURE);
        } else if ( pids[0] == 0 ) {
            continue;
        }

        pids[1] = fork();
        if ( pids[1] == -1 ) {
            perror("fork failed");
            exit(EXIT_FAILURE);
        } else if ( pids[1] == 0 ) {
            continue;
        }

        for ( int i = 0; i < 2; i++ ) {
            if ( waitpid(pids[i], NULL, 0) == -1 ) {
                perror("waitpid failed");
                exit(EXIT_FAILURE);
            }
        }

        break;
    }

    printf("level %d: parent %ld, child %ld\n", level, (long) getppid(), (long) getpid());

    return 0;
}

带输出，显示完全平衡的二叉树：

me@mac:$ ./fork | sort
level 1: parent 62427, child 73103
level 2: parent 73103, child 73105
level 2: parent 73103, child 73106
level 3: parent 73105, child 73107
level 3: parent 73105, child 73109
level 3: parent 73106, child 73108
level 3: parent 73106, child 73111
level 4: parent 73107, child 73110
level 4: parent 73107, child 73114
level 4: parent 73108, child 73112
level 4: parent 73108, child 73116
level 4: parent 73109, child 73113
level 4: parent 73109, child 73117
level 4: parent 73111, child 73115
level 4: parent 73111, child 73118

我正在尝试在 C 中制作 fork() 进程调用图

I'm trying to make a diagram of fork() process calls in C

c

fork

pid