多少进程与这些fork（）的语句创建？ [英] How many processes are created with these fork() statements?

问题描述

我相信，这将创建24处理;不过，我需要核实。这些问题常常难倒我了。感谢您的帮助！

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
＃包括LT＆;＆unistd.h中GT;INT主要（无效）
{
  将为pid_t PID =叉（）;
  PID =叉（）;
  PID =叉（）;
  如果（PID == 0）
  {
    叉子（）;
  }
  叉子（）;
  返回0;
}
 

解决方案

这是很容易通过这个道理。在叉调用创建每次它执行时的附加处理。调用返回 0 新（子）进程，并在原始（父）过程中的孩子（不为零）的进程ID。

 将为pid_t PID =叉（）; //叉＃1
PID =叉（）; //叉＃2
PID =叉（）; //叉＃3
如果（PID == 0）
{
  叉子（）; //叉＃4
}
叉子（）; //叉＃5
 

1. 叉＃1创建一个额外的进程。你现在有两个进程。


2. 叉＃2由两个进程执行时，创建两个过程，总共四个


3. 叉＃3由四个进程执行时，创建四个过程，总共八个。其中一半有 PID == 0 半有 PID！= 0


4. 叉＃4是由叉＃3中创建的过程中执行的一半（所以，他们四人）。这将创建四个额外的进程。您现在有十二流程。


5. 叉＃5被其他进程的所有十二个执行，创造再多得十几个过程;你现在有24。

I believe that this creates 24 processes; however, I need verification. These questions often stump me. Thanks for the help!

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

int main(void)
{
  pid_t pid = fork();
  pid = fork();
  pid = fork();
  if (pid == 0)
  {
    fork();
  }
  fork();
  return 0;
}

解决方案

It's fairly easy to reason through this. The fork call creates an additional process every time that it's executed. The call returns 0 in the new (child) process and the process id of the child (not zero) in the original (parent) process.

pid_t pid = fork();  // fork #1
pid = fork();        // fork #2
pid = fork();        // fork #3
if (pid == 0)
{
  fork();            // fork #4
}
fork();              // fork #5

1. Fork #1 creates an additional processes. You now have two processes.
2. Fork #2 is executed by two processes, creating two processes, for a total of four.
3. Fork #3 is executed by four processes, creating four processes, for a total of eight. Half of those have pid==0 and half have pid != 0
4. Fork #4 is executed by half of the processes created by fork #3 (so, four of them). This creates four additional processes. You now have twelve processes.
5. Fork #5 is executed by all twelve of the remaining processes, creating twelve more processes; you now have twenty-four.

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