使用fork时如何映射内存? [英] How the memory is mapped when fork is used?

问题描述

我是"fork()"的新手，我到处都读到，当fork()被调用时，当前(调用)进程的精确副本已启动.有两个不同的内存位置分配给它们的变量和函数.

i am new to "fork()",I read everywhere that when a fork() is called an exact copy of current (calling) process is started.Now when I run following code ,there should be two different processes, having two different memory locations assigned to their vars and functions.

#include<stdio.h>
int i=10;
int pid;
 int main(){
  if((pid=fork())==0){
    i++;//somewhere I read that separate memory space for child is created when write is needed
    printf("parent address= %p\n",&i);// this should return the address from parent's memory space
  }else{
    i++;
    i++;
    printf("child address= %p\n",&i);// this should return the address of child's memory space 
  }
  wait(0);
  return(0);
 }

Why The output looks like:: 
child address::804a01c 
parent address::804a01c

为什么父母和孩子的地址都相同?

Why both the address are the same for parent as well as child?

推荐答案

为它们的var和函数分配了两个不同的存储位置.

having two different memory locations assigned to their vars and functions.

没有； Linux实现虚拟内存，这意味着每个进程都有自己的完整地址空间.结果，在fork之后，两个进程的内存对象副本都将看到相同的地址.

Nope; Linux implements virtual memory, meaning that each process has its own complete address space. As a result, after a fork, both processes see the same addresses for their copies of in-memory objects.

(顺便说一句:VM还会使代码在物理内存中的进程之间共享，并且所有数据将仅写入时复制.)

(As an aside: VM also causes code to be shared between the process in physical memory, and all data will only be copied-on-write.)

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