CS202: HW 5: More concurrency

These problems should be done on your own. You get credit if you turn in something, and we are not checking the correctness. So these exercises exist purely to reinforce the material, not to evaluate you.

Getting help from AI (besides being ruled out by the course policies) will actually hurt you, since the point of these questions is for you to gain the practice and experience of working through the problems.

You will need that general skill (of absorbing something by practicing it) in life, and in this semester you will need the specific problem-solving skills that are emphasized in these homeworks.

Sleeping barber

This is a potentially useful (if convoluted) example to exercise your understanding of how to use mutexes and conditional variables. It is a well-known concurrency problem. The writeup and solution are due to Mike Dahlin (who used to be on the faculty at The University of Texas at Austin). He asked this question on a midterm once.

Work through the problem on your own; we will post the solution next week.

A shop has a barber, a barber chair, and a waiting room with NCHAIRS chairs. If there are no customers present, the barber sits in the chair and falls asleep. When a customer arrives, the customer wakes the sleeping barber. If an additional customer arrives while the barber is cutting hair, the customer sits in a waiting room chair if one is available. If no chairs are available, the customer leaves the shop. When the barber finishes cutting a customer’s hair, the barber tells the customer to leave; then, if there are any customers in the waiting room, the barber announces that the next customer can sit down. Customers in the waiting room get their hair cut in FIFO order.

The barber shop can be modeled as 2 shared objects:

• A BarberChair, with the methods napInChair(), wakeBarber(), sitInChair(), cutHair(), and tellCustomerDone(). The BarberChair must have a state variable with the following states: EMPTY, BARBER_IN_CHAIR, LONG_HAIR_CUSTOMER_IN_CHAIR, SHORT_HAIR_CUSTOMER_IN_CHAIR.

  • Note that neither a customer or barber should sit down until the previous customer is out of the chair (state == EMPTY).

  • Note that cutHair() must not return until the customer is sitting in the chair (LONG_HAIR_CUSTOMER_IN_CHAIR).

  • Note that a customer should not get out of the chair (that is, return from sitInChair()) until the customer’s hair is cut (SHORT_HAIR_CUSTOMER_IN_CHAIR).

  • The barber should get in the chair (BARBER_IN_CHAIR) only if no customers are waiting.

  • You may need additional state variables.

• A WaitingRoom, with the methods enter() and callNextCustomer().

  • enter() returns WR_FULL if the waiting room is full or (immediately or eventually) returns MY_TURN when it is the caller’s turn to get their hair cut

  • callNextCustomer() returns WR_BUSY or WR_EMPTY depending on if there is a customer in the waiting room or not. Customers are served in FIFO order.

Thus, each customer thread executes the code:

Customer(WaitingRoom *wr, BarberChair *bc)
{
    status = wr->enter();
    if (status == WR_FULL) {
        return;
    }
    bc->wakeBarber();
    bc->sitInChair();   // Wait for chair to be EMPTY
                        // Make state LONG_HAIR_CUSTOMER_IN_CHAIR
                        // Wait until SHORT_HAIR_CUSTOMER_IN_CHAIR
                        // then make chair EMPTY and return
    return;
}

The barber thread executes the code:

Barber(WaitingRoom *wr, BarberChair *bc)
{
    while (1) { // A barber’s work is never done
        status = wr->callNextCustomer();
        if (status == WR_EMPTY) {
            bc->napInChair(); // Set state to BARBER_IN_CHAIR; return with state EMPTY
        }
        bc->cutHair(); // Block until LONG_HAIR_CUSTOMER_IN_CHAIR;
                       // Return with SHORT_HAIR_CUSTOMER_IN_CHAIR
        bc->tellCustomerDone(); // Return when EMPTY
    }
}

Deadlock

// assume all the variables are initialized correctly
double balance[2]; // 0 for alice, 1 for bob
smutex_t mtx[2];    // 0 for alice, 1 for bob

bool transfer(int from, int to, double trans) {
  smutex_lock(&mtx[from]);
  smutex_lock(&mtx[to]);

  bool result = false;
  if (balance[from] > trans) {
    balance[from] = balance[from] - trans;
    balance[to] = balance[to] + trans;
    result = true;
  }

  smutex_unlock(&mtx[to]);
  smutex_unlock(&mtx[from]);
  return result;
}

More Deadlock

You are programming an application that has a very large array as a data structure, and multiple threads of control concurrently operate on the data structure. Performing a given operation on the array requires a thread to modify several array items at once, but the application logic requires all of the modifications associated with a given operation to happen atomically. Also, you cannot predict at compile time which items your code will be accessing together.

You approach the problem as follows. (1) For maximum concurrency, you create a lock for each item in the array. (2) When a thread performs an operation, it: (a) acquires all of the locks for the items that the operation needs to modify, then (b) modifies the array items, and then (c) releases all of the locks. This approach provides the required atomicity, but it creates the risk of deadlock.

A common bug in C coding

Consider the following code:

#include <stdlib.h>
#include <stdio.h>

// A box. Each box has an ID and a pointer to the box that resides inside
// of it. If the box has nothing inside of it, inner_box should be equal
// to NULL.
struct box {
    int id; 
    struct box *inner_box;
};

// Insert box: places the box "inner" inside of the box "outer".
// Since "outer" is being modified, we pass a pointer to "outer".
// Since "inner" is not being modified, we pass in "inner" directly.
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;
}

// Print box: prints a box and the box inside of it. This function
// is recursive and will end once a box is empty.
void print_box(struct box* first, int level) {
    int i;
    if (!first) 
        return;

    for (i=0; i < level; ++i) {
        printf("- ");
    }       
    printf("id: %d\n", first->id);
    print_box(first->inner_box, level+1);
}

int main() {
    // Create three boxes.
    struct box box1 = { .id = 37, .inner_box = NULL };
    struct box box2 = { .id = 12, .inner_box = NULL };
    struct box box3 = { .id = 19, .inner_box = NULL };

    // The box ordering should be box1 -> box2 -> box3
    insert_box(&box1, box2);
    insert_box(&box2, box3);
    
    // Print the boxes starting from the outside box.
    print_box(&box1, 0);

    return 0;
}

Reader Writer Spinlock (Challenge)

  struct sharedlock {
    int value; // when the lock is created, value is initialized to 0
  };

  // we are giving you the code for the first of the four functions:
  void reader_acquire(struct sharedlock* lock) {
    int curr_val;
    while (1) {

      // spin while a writer owns the lock
      while ((curr_val = lock->value) == -1) {}

      assert(curr_val >= 0);

      // try to atomically increment the count, based on our best
      // guess of how many readers there had been. if we were
      // wrong, keep looping. if we got it right, then we
      // succeeded in incrementing the count atomically, and we
      // can proceed.
      if (cmpxchg_val(&lock->value, curr_val, curr_val + 1) == curr_val)
        break;
    }
    // lock->value now contains curr_val + 1
  }

Appendix: Implementation of atomic primitives

Here is pseudocode and actual x86 assembly code for the implementations of the atomic primitives in the reader-writer spinlock question.

cmpxchg_val()

  /* pseudocode */
  int cmpxchg_val(int* addr, int oldval, int newval) {
    LOCK: // remember, this is pseudocode
    int was = *addr;
    if (*addr == oldval)
      *addr = newval;
    return was;
  }

  /* inline assembly */
  int cmpxchg_val(int* addr, int oldval, int newval) {
      int was;
      asm volatile("lock cmpxchg %3, %0"
        : "+m" (*addr), "=a" (was)
        : "a" (oldval), "r" (newval)
        : "cc");
       return was;
  }

atomic_decrement()

  /* pseudocode */
  void atomic_decrement(int* arg) {
    LOCK: // remember, this is pseudocode
    *arg = *arg - 1;
  }

  /* inline assembly */
  void atomic_decrement(int* arg) {
    asm volatile("lock decl %0" : "+m" (*arg) : "m" (arg));
  }

Handing in the homework

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