GATE CS 2021 Set 2 — Question 52

MSQ+2 / -0MediumProcesses & ThreadsOperating SystemMutex & SpinlocksProcess Synchronization

Operating System → Processes & Threads → Mutex & Spinlocks

Last updated 5 September 2026

Question

Consider the following multi-threaded code segment (in a mix of C and pseudo-code), invoked by two processes P1 and P2, and each of the processes spawns two threads T1 and T2: int x = 0; // global Lock L1; // global main() { create a thread to execute foo(); // Thread T1 create a thread to execute foo(); // Thread T2 wait for the two threads to finish execution; print (x); } foo() { int y = 0; Acquire L1; x = x + 1; y = y + 1; Release L1; print (y); } Which of the following statement(s) is/are correct?

Correct answer (A) Both P1 and P2 will print the value of x as 2.; (D) Both T1 and T2, in both the processes, will print the value of y as 1. Solution Step-by-step analysis: 1. Process Isolation : Processes P1 and P2 have separate address spaces. Global variables like x and L1 are local to each process. They are not shared between P1 and P2. 2. Process P1 : It spawns two threads T1 and T2. These threads share the same address space, including x and L1 . Both threads call foo() . Inside foo(), the critical section x = x + 1 is protected by Acquire L1 and Release L1 . Thus, x will be incremented exactly twice (once by T1, once by T2). The final value of x printed by P1 will be 2. 3. Process P2 : Similarly, P2 has its own x and L1 . Its threads T1 and T2 will increment its own x twice. The final value of x printed by P2 will also be 2. Thus, statement (A) is correct and (B) is incorrect. 4. Variable y : The variable y is declared inside foo(), making it a local variable. In multi-threaded environments, each thread has its own stack, and thus its own copy of local variables. 5. Thread Execution : For every thread (T1 and T2 in both P1 and P2), y is initialized to 0, incremented to 1, and then printed. No thread shares y with another. Therefore, every thread will print y as 1. Thus, statement (D) is correct and (C) is incorrect. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (A) Both P1 and P2 will print the value of x as 2.; (D) Both T1 and T2, in both the processes, will print the value of y as 1. Solution Step-by-step analysis: 1. Process Isolation : Processes P1 and P2 have separate address spaces. Global variables like x and L1 are local to each process. They are not shared between P1 and P2. 2. Process P1 : It spawns two threads T1 and T2. These threads share the same address space, including x and L1 . Both threads call foo() . Inside foo(), the critical section x = x + 1 is protected by Acquire L1 and Release L1 . Thus, x will be incremented exactly twice (once by T1, once by T2). The final value of x printed by P1 will be 2. 3. Process P2 : Similarly, P2 has its own x and L1 . Its threads T1 and T2 will increment its own x twice. The final value of x printed by P2 will also be 2. Thus, statement (A) is correct and (B) is incorrect. 4. Variable y : The variable y is declared inside foo(), making it a local variable. In multi-threaded environments, each thread has its own stack, and thus its own copy of local variables. 5. Thread Execution : For every thread (T1 and T2 in both P1 and P2), y is initialized to 0, incremented to 1, and then printed. No thread shares y with another. Therefore, every thread will print y as 1. Thus, statement (D) is correct and (C) is incorrect.

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