The enter_CS() and leave_CS() functions to implement critical section ...
The above solution is a simple test-and-set solution that makes sure that deadlock doesn’t occur, but it doesn’t use any queue to avoid starvation or to have FIFO order.
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The enter_CS() and leave_CS() functions to implement critical section ...
Statement I: The above solution to CS problem is deadlock-free
The given solution uses the test-and-set instruction to implement the critical section. The test-and-set instruction atomically sets a memory location to a specific value and returns the previous value. In this case, the memory location X is initially set to 0.
To enter the critical section, a process must successfully execute the test-and-set instruction on X. If the value of X is 0 (i.e., not set by any other process), the test-and-set instruction will set X to 1 and return 0, allowing the process to enter the critical section. If the value of X is 1 (i.e., already set by another process), the test-and-set instruction will not modify X and return 1, causing the process to spin in a while loop until it can successfully enter the critical section.
Since the test-and-set instruction guarantees mutual exclusion, only one process can successfully set X to 1 and enter the critical section at a time. This ensures that there will be no deadlock in the system. Deadlock occurs when multiple processes are waiting indefinitely for a resource that is held by another process, but in this case, the processes will eventually be able to enter the critical section.
Therefore, Statement I is TRUE.
Statement II: The solution is starvation-free
Starvation occurs when a process is prevented from making progress indefinitely due to resource allocation policies. In this case, the test-and-set instruction ensures that only one process can enter the critical section at a time.
Since the processes spin in a while loop until they can successfully enter the critical section, there is no guarantee of fairness or ordering among the processes. It is possible for a process to be continuously preempted by other processes, causing it to wait indefinitely and potentially starve.
Therefore, Statement II is FALSE.
Statement III: The processes enter CS in FIFO order
The given solution does not enforce any explicit ordering or fairness among the processes. The processes simply spin in a while loop until they can successfully enter the critical section. The order in which the processes enter the critical section depends on various factors, such as the timing of their test-and-set instructions and the scheduling policy of the system.
Therefore, Statement III is FALSE.
Statement IV: More than one process can enter CS at the same time
Since the test-and-set instruction guarantees mutual exclusion, only one process can successfully set X to 1 and enter the critical section at a time. Other processes will spin in a while loop until they can successfully enter the critical section. Therefore, at any given time, only one process can be in the critical section.
Therefore, Statement IV is FALSE.
In conclusion, the only true statement is Statement I: The above solution to the CS problem is deadlock-free.
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