Question

Fetch_And_Add(X,i) is an atomic Read-Modify-Write instruction that reads the value of memory location X, increments it by the value i, and returns the old value of X, it is used in the pseudocode shown below to implement a busy-wait lock. L is unsigned integer shared variable initialized to 0. The value of 0 corresponds to lock being available, while any non-zero value corresponds to the lock being not available.

AcquireLock (L) {
while (Fetch_And_Add(L,1))
L = 1;
}
Release Lock (L){
L=0;
}
This implementation

• A. fails as L can take on a non-zero value when the lock is actually available
• B. works correctly but may starve some processes
• C. fails as L can overflow
• D. works correctly without starvation

Answer 1

The correct option is A fails as L can take on a non-zero value when the lock is actually available

Assume $P_1$ executes until while condition and preempts before executing L = 1. Now $P_2$ executes all statements, hence L = 0. Then $P_1$ without checking L it makes L = 1 by executing the statement where it was preempted.
$\therefore$ It takes a non-zero value (L=1) when the lock is actually available (L=0)