Execution when using fork in a program - linux

Question

I was reading about fork() when I tried out the following program. I could not understand the output of the following command but I could figure out what it does if I remove the second fork() call. Please explain me the flow of the following program.
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <error.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
using namespace std;
int main(int argc, char const *argv[])
{
pid_t pid;
int returnvalue;
pid = fork();
pid = fork();
if (!pid) cout<< "In the child"<<endl;
else cout<< "In parent"<<endl;
cout<< (&returnvalue)<<endl;
switch(pid)
{
case -1:
perror("fork");
return 0;
case 0:
cout<< "Child Process with pid: " <<getpid()<<endl;
cout<< "Parent's pid is: "<<getppid()<<endl;
cout<< "Exiting"<<endl;
returnvalue=2;
return returnvalue;
default:
cout<< "Parent process with pid: "<<getpid()<<endl;
cout<< "Child's pid: "<<pid<<endl;
cout<< "Waiting for child to exit"<<endl;
wait(&returnvalue);
cout<< "Child's exit status: "<<WEXITSTATUS(returnvalue)<<endl;
cout<< "Exiting!"<<endl;
}
return 0;
}
Why does it print "In parent" and "In the child" twice each?
Also, I read that every child process gets its own copy of variables. Shouldn't address of "returnvalue" be printed different?
Output:
In parent
0x7fff2d536428
Parent process with pid: 5487
Child's pid: 5489
Waiting for child to exit
In the child
0x7fff2d536428
Child Process with pid: 5489
Parent's pid is: 5487
Exiting
Child's exit status: 2
Exiting!
In parent
0x7fff2d536428
Parent process with pid: 5488
Child's pid: 5490
Waiting for child to exit
In the child
0x7fff2d536428
Child Process with pid: 5490
Parent's pid is: 5488
Exiting
Child's exit status: 2
Exiting!

Answer 1

Well you do have 2 calls to fork() will will result in 4 processes.
1st fork in P1 --> new process P2
2nd fork in P1 --> new process P3
2nd fork in P2 --> new process P4
the pid will be different based on the result following the 2nd fork() for sake of the explanation, I'll just say that P1 is the parent and remains the parent, P2 is a child of P1, but is a parent of P4, so its pid will be non zero, both P3 and P4 will have pid equal to 0. All 4 processes will enter the switch statement with a pid that will classify them as a parent or child, so since 2 have pid = 0 and 2 have pid != 0, 2 will be reported as parents and 2 will be reported as children.
P1 pid != 0 (classify parent)
P2 pid != 0 (classify parent)
P3 pid == 0 (classify child)
P4 pid == 0 (classify child)
It is true that P1 is created before P2 is created before P3 is created before P4, however when they enter into the switch statement and when they print out their messages is controlled by the scheduler. Consider the possibility that creating a process takes a lot of time (more then just printing stuff). So P1 creates P2 and then turns around and creates P3 meanwhile P2 is busy creating P4. P3 get created and prints stuff while P2 is still stuck creating P4.