GATE2025_CS1_Q29

Suppose in a multiprogramming environment, the following C program segment is executed. A process goes into I/O queue whenever an I/O related operation is performed. Assume that there will always be a context switch whenever a process requests for an I/O, and also whenever the process returns from an I/O. The number of times the process will enter the ready queue during its lifetime (not counting the time the process enters the ready queue when it is run initially) is ____. (Answer in integer)

int main() { int x=0, i=0; scanf("%d", &x); for(i=0; i<20; i++) { x = x+20; printf("%d\n",x); } return 0; }

Your Answer (Integer): ________

✅ Final Answer:

The correct answer is 21.

🔹 Short Answer:

The process enters the ready queue each time it completes an I/O operation. The `scanf` function is one I/O operation. The `printf` function inside the `for` loop is called 20 times, resulting in 20 more I/O operations. Therefore, the total number of times the process enters the ready queue (excluding the initial entry) is 1 (for `scanf`) + 20 (for `printf`) = 21.

🔸 Long Answer:

This question tests the understanding of process state transitions in an operating system, specifically in response to I/O operations.

Process State Transitions

A process typically moves between three basic states: Running, Ready, and Blocked (or Waiting).

• A process in the Running state is currently being executed by the CPU.
• A process in the Ready state is waiting to be assigned to the CPU.
• A process in the Blocked state is waiting for some event to occur, such as the completion of an I/O operation.

The question specifies that whenever an I/O operation is requested, the process moves to a waiting state, and upon I/O completion, it moves to the ready state. This is the key transition we need to count.

Code Execution Analysis

Let's trace the execution and count the transitions to the Ready queue, remembering to ignore the very first one.

1. Initial Run: The process is created and enters the Ready queue for the first time. The question asks us to not count this. It is then scheduled and moves to the Running state.
2. scanf("%d", &x);:
   • This is a blocking I/O operation.
   • The process transitions from Running → Blocked.
   • When the user provides input and presses Enter, the I/O operation completes.
   • The process transitions from Blocked → Ready. (Count = 1)
3. for(i=0; i<20; i++):
   • This loop will execute 20 times (for i = 0, 1, ..., 19).
   • Inside the loop, x = x+20; is a CPU operation and does not cause a state change from Running.
   • printf("%d\n",x);: This is also treated as a blocking I/O operation. For each of the 20 iterations:
     • The process transitions from Running → Blocked.
     • When the output operation completes, the process transitions from Blocked → Ready.
   • This loop contributes 20 entries to the Ready queue.

Final Calculation

Total entries into the Ready Queue (after the initial run) = (Entries from scanf) + (Entries from printf loop)

Total = 1 + 20 = 21.

Bibliography:
- Silberschatz, A., Galvin, P. B., & Gagne, G. (2018). Operating System Concepts. Wiley. (Chapter 3: Processes).
- Stallings, W. (2017). Operating Systems: Internals and Design Principles. Pearson. (Chapter 3: Process Description and Control).

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