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?
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.)