子进程中的变量修改

问题描述

我正在研究Bryant和O'Hallaron的计算机系统，程序员的观点.练习8.16要求提供如下程序的输出(我更改了程序，因为他们使用了可以在其网站上下载的头文件):

I am working on Bryant and O'Hallaron's Computer Systems, A Programmer's Perspective. Exercise 8.16 asks for the output of a program like (I changed it because they use a header file you can download on their website):

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <errno.h>
#include <unistd.h>
#include <string.h>
int counter = 1;

int main()
{
    if (fork() == 0){
        counter--;
        exit(0);
    }

    else{
        Wait(NULL);
        printf("counter = %d\n", ++counter);
    }
    exit(0);
}

我回答"counter = 1"，因为父进程等待其子进程终止，然后递增counter.但是孩子首先要减少它.但是，当我测试该程序时，我发现正确的答案是"counter = 2".子进程和父进程中的变量"counter"是否不同?如果没有，那么答案为何2?

I answered "counter = 1" because the parent process waits for its children to terminate and then increments counter. But the child first decrements it. However, when I tested the program, I found that the correct answer was "counter = 2". Is the variable "counter" different in the child and in the parent process? If not, then why is the answer 2?

推荐答案

您的父进程以 1 处的 counter 开始.

Your parent process starts with counter at 1.

然后等待分叉的子进程完成.

Then it waits for the forked child process to finish.

(分叉进程中发生的任何事情都不会影响父级 counter 的版本.每个进程都有自己的内存空间；不共享任何变量.)

(Whatever happens in the forked process, does not affect the parent's version of counter. Each process has its own memory space; no variables are shared.)

然后，最后 printf()语句首先使用 ++ 运算符将 counter 递增，这使 counter 获取值 2 .

And, finally the printf() statement first increments counter with the ++ operator, this makes counter get the value 2.

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