The first three problems in this Exercise refer to a critical section of the form
lock(lk);
operation
unlock(lk);
where the “operation” updates the shared variable shvar using the local (nonshared) variable x as follows:

If we want to atomically perform the same operation on two shared variables (e.g., shvar1 and shvar2) in the same critical section, we can do this easily using the approach from 2.29.1 (simply put both updates between the lock operation and the corresponding unlock operation). Explain why we cannot do this using the approach from 2.29.2. i.e., why we cannot use ll/sc to access both shared variables in a way that guarantees that both updates are executed together as a single atomic operation.

Problem 2.29.2

The first three problems in this Exercise refer to a critical section of the form
lock(lk);
operation
unlock(lk);
where the “operation” updates the shared variable shvar using the local (nonshared) variable x as follows:

Repeat problem 2.29.1, but this time use ll/sc to perform an atomic update of the shvar variable directly, without using lock() and unlock(). Note that in this problem there is no variable lk.

problem 2.29.1

Write the MIPS assembly code for this critical section, assuming that the address of the lk variable is in $a0, the address of the shvar variable is in $a1, and the value of variable x is in $a2. Your critical section should not contain any function calls, i.e., you should include the MIPS instructions for lock(), unlock(), max(), and min() operations. Use ll/sc instructions to implement the lock() operation, and the unlock() operation is simply an ordinary store instruction.

Transcribed Image Text:

a. shvar=max(shvar,x); b. if(shvar>0) shvar=max(shvar,x); Operation