#OS process synchronization

Question

What happens if we run the following program?
int main()
{
int sem;
sem = screate(1); //Semaphore is created and is initialized to 1
fork ();
Wait(sem);
printf(―Gate ―);
Signal(sem);
return 0;
}
(A) Starvation occurs
(B) Deadlock occurs
(C) Both processes enter critical section simultaneously
(D) ―Gate Gate―will be the output

Answer 1

Actually when fork() executed,

child process created, therefore now two processes running from the next point point in the program

The execution of child processes is independent on Parent process therefore interleaved execution may happened

by using semaphore we are restricting that interleaving process therefore either Parent or Child process can enter into CS but at a time only one process.

Therefore Option D is correct.

---

For observing the difference,

fork();

Wait(sem);

printf(―Gate ―);

Signal(sem);

replace with 

int t=fork();

Wait(sem);

if(t==0)
{
    printf( ― CHILD LINE1 ― );
    printf( ― CHILD LINE2 ― );
}

else
{
    printf( ― PARENT LINE1 ― );
    printf( ― PARENT LINE2 ― );
}

Signal(sem);

then o/p is ONE OF THESE

1) ― PARENT LINE1 ― ― PARENT LINE2 ― ― CHILD LINE1 ―― CHILD LINE2 ― 

2)  ― CHILD LINE1 ―― CHILD LINE2 ― ― PARENT LINE1 ― ― PARENT LINE2 ―

 

what happend if we doesn't use semaphores in the above example?

one of the possible outcome may be 

1)  ― PARENT LINE1 ― ― CHILD LINE1 ― ― CHILD LINE2 ―  ― PARENT LINE2 ―

2)   ― PARENT LINE1 ― ― CHILD LINE1 ― ― PARENT LINE2 ― ― CHILD LINE2 ― 

Comments

Parent and child both will have their own copy of sem variable or shared ?

Instead of semaphore if normal local variable then both parent and child will have different copy of variable ?

---

@rajankakaniya both parent and child will have shared copy of sem variable as sem is acting here as synchronisation construct. If instead of semaphore local variables are present then both parent and child will have different copy of variables. 

Answer 2

C. as there is a fork so two process will be made parent and child  and they both have a same virtual address and different physical address so they both will be having same sem=1

so after getting exectued wait (Sem) parent prempted and now child come into running state and do wait(Sem) so both the process enter into <CS>

some of you would be having that why both will be having sem=1 as sem is local variable so both parent and child will be having sem=1

Comments

generally we assume semaphores are kept under shared memory !

https://stackoverflow.com/questions/6847973/do-forked-child-processes-use-the-same-semaphore