מערכות הפעלה

1. It will create two process one parent P (has process ID of child process) and other is child C1 (process ID = 0). 2. When condition is true parent P executes if statement and child C1 executes else statement and print 3. Parent P checks next if statement and create two process (one parent P and child C2). In if statement we are using not operator (i.e, !), it executes for child process C2 and parent P executes else part and print value 2. Child C2 further creates two new processes (one parent C2 and other is child C3).

1. Fork will create two process one parent P (has process id of new child) and other one is child C1 (process id=0). 2. In if statement we are using AND operator (i.e, &&) and in this case if first condition is false then it will not evaluate second condition and print 2. Parent process P check for second condition and create two new processes (one parent P and other is child C2). In second condition we are using NOT operator which return true for child process C2 and it executes inner if statement. 3. Child C2 again create two new processes (one parent C2 and child C3) and we are using OR operator (i.e, ||) which evaluate second condition when first condition is false. Parent C2 execute if part and create two new processes (one parent C2 and child C4) whereas child C3 check for second condition and create two new processes (one parent C3 and child C5). 4. Parent C3 enters in if part and further create two new processes (one parent C3 and child C6).

The number of times ‘hello’ is printed is equal to number of process created. Total Number of Processes = 2^n, where n is number of fork system calls. So here n = 3, 2^3 = 8

fork() returns 0 in child process and process ID of child process in parent process. In Child (x), a = a + 5 In Parent (u), a = a – 5; Therefore x = u + 10. The physical addresses of ‘a’ in parent and child must be different. But our program accesses virtual addresses (assuming we are running on an OS that uses virtual memory). The child process gets an exact copy of parent process and virtual address of ‘a’ doesn’t change in child process. Therefore, we get same addresses in both parent and child. But in python3 v and y will not be equal.

At time 0, P1 is the only process, P1 runs for 15 time units. At time 15, P2 arrives, but P1 has the shortest remaining time. So P1 continues for 5 more time units. At time 20, P2 is the only process. So it runs for 10 time units At time 30, P3 is the shortest remaining time process. So it runs for 10 time units At time 40, P2 runs as it is the only process. P2 runs for 5 time units. At time 45, P3 arrives, but P2 has the shortest remaining time. So P2 continues for 10 more time units. P2 completes its ececution at time 55 Total waiting time for P2 = Complition time - (Arrival time + Execution time) = 55 - (15 + 25) = 15

The above synchronization constructs don’t prevent deadlock. When both wants1 and wants2 become true, both P1 and P2 stuck forever in their while loops waiting for each other to finish.