HW5 Solutions

1. 

   A B C  

   B A C  

    ASSUMPTION: A does not wait, sleep, or block once it has acquired
    the mutex.

    One way to answer this is to observe that the mutex is the only
    scheduling constraint that we are explicitly given. For this
    constraint, there are two cases: either A acquires the mutex first,
    or C does. 
    
    In the first case, A, being high priority, executes to completion.
    Then B, Then C.  (Note that if A must go to sleep in the critical
    section, then we get the B A C output.)
   
    In the second case, C acquires the mutex before A (probably because
    A's "do something" took a while). But C is low priority, so even
    though C got to the mutex first, B was scheduled ahead of C, i.e., B
    managed to run first. So B finishes. When C releases the mutex, A
    can acquire it, and since A is high priority, A runs to completion,
    printing A. Then C finally finishes. 

     [NOTE: you might wonder whether 
        B C A
     is a possible output. The reason not is that
     even if C acquires the mutex first, at the instant that C releases
     it, A will run to completion before C's "do something" block.]

2.
        
    void reader_release(sharedlock* lock)
    {
        atomic_decrement(&lock->value);
    }

    void writer_acquire(sharedlock* lock)
    {
        /*
         * a common error here is not spinning, i.e., not including
         * the while() around the cmpxchg. note that spinning is what keeps
         * the writer from entering the critical section before it's
         * supposed to
         */

        while (cmpxchg(&lock->value, 0, -1) != 0) {}
    }

    void writer_release(sharedlock* lock)
    {
        xchg_val(&lock->value, 0); /* xchg_val is from class notes */
        /*
         * two other options:
         * 1.  cmpxchg(&lock->value, -1, 0);
         * 2.  lock->value = 0;
         * option 2. works because the question said to
         * assume sequential consistency. Without sequential
         * consistency, option 2. might not work because it might
         * not guarantee that all of the instructions before or after it
         * would appear in program order to the other CPUs in the machine.
         */
    }

3.

3.1: The code is designed to place box2 inside of box1, and box3 inside of box2. The expected output is:
37
- 12
- - 19


3.2: The program's behavior is undefined, as print_box() will deference
an undefined pointer. Possible outputs include an infinite loop,
printing of garbage values, or a segmentation fault.


3.3: When a function is invoked with a call-by-value parameter, the
parameter is copied and the function executes with a local copy. In this
case, insert_box() makes a copy of inner; insert_box() then sets
outer->inner_box to the address of this copy. But where does the copy
live? It lives on the _stack_ and will go out of scope when the function
returns. Thus, when the function returns outer->inner_box points to
invalid memory (concretely, it is pointing to a location on the stack
that will be used for something else). 

3.4: Change insert_box to use a pointer for inner:
void insert_box(struct box* outer, struct box* inner) {
    printf("insert box: placing id %d inside id %d\n", inner->id, outer->id);
    outer->inner_box = inner;
}

Also, change main() to correctly invoke the modified function:
int main() {
    ...
    insert_box(&box1, &box2);
    insert_box(&box2, &box3);
    ...
}