Question

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

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

Answer 1

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

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.

(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.)