Question 1:

  In this problem, you will complete a function that converts "float" to
  "int" without explicit use of a conversion operator. 

    a. u & (1 << 31)
    b. (u & 0x7f800000) >> 23
    c. u & 0x007fffff
    d. exp == 0
    e. exp == 255
    f. frac != 0
    g. shift = 23 - (exp - 127);



Question 2:

  Given the assembly for the function mystery1(), fill in the corresponding
  function in C.

    int mystery1(int A[], int n)
    {
      int i;

      int mystery = 0;

      for (i=0; i<n; i++)
      {
        mystery += A[i];
      }

      return mystery;
    }

  
  A. What would the following function call return?
     
        x = 2 

  B. What is the mystery2 function computing?

        Greatest Common Denominator (GCD)



Question 3:

   Part I:
                          (0x804961c/0x37)
                           /            \
               (0x8049604/0x1f)     (0x8049628/0x3c)
                 /           \                  \
         (0x80495f8/0xc) (0x8049610/0x22)     (0x8049634/0x4e)

   Part II:

       <Address>   <Value>
       0xbffff800   0x4e
       0xbffff7fc   0x804961c
       0xbffff7f8   rtn_addr
       0xbffff7f4   0xbffff880
       0xbffff7f0   0x4e
       0xbffff7ec   0x8049628
       0xbffff7e8   rtn_addr
       0xbffff7e4   0xbffff7f4
       0xbffff7e0   0x4e
       0xbffff7dc   0x8049634
       0xbffff7d8   rtn_addr
       0xbffff7d4   0xbffff7e4



Question 4: 

  A. 1/12
  B. 1/6
  C. 1/3
  D. 1/6



Question 5:

  1. VPO - [8:0]    TLBI - [9]
     VPN - [17:9]   TLBT - [17:10]

  2. PPO - [8:0]    CO   - [1:0]   CT - [11:4]
     PPN - [11:9]   CI   - [3:2]

  A. 01 1010 1001 1111 0100

  B. VPN         - 0xD4
     TLBI        - 0
     TLBT        - 0x6A
     TLB Hit?    - Y
     Page Fault? - N
     PPN         - 0x3

  C. 0111 1111 0100

  D. BO          - 0
     CI          - 1
     CT          - 0x7F
     Cache Hit?  - Y
     Byte        - 0xFF



Question 6:

  Part 1:
    A. 127 B / 2048 B
    B. internal fragmentation is the space wasted because the request
       is smaller than the payload.  external fragmentation is the space
       wasted because the unallocated sections of the heap are not
       contiguous, and thus some allocation requests cannot be fulfilled

  Part 2:
    A. O(N-M)
    B. O(N)
    C. O(N-M)

  Part 3:
    A. O(1)
    B. O(N-M)
    C. lazy
    D. immediate
    E. lazy or doesn't matter



Question 7:

  Part 1:

    Boxes 1,2 and 4 checked.

  Part 2:

    #include <stdio.h>
    #include <pthread.h>

    #define NTHREADS 20

    pthread_mutex mutex;    <============================ Add this


    void *thread(void *vargp)
    {

      static int cnt = 0;

      pthread_mutex_lock(&mutex); <====================== Add this

      cnt++;

      printf("%d\n", cnt);

      pthread_mutex_unlock(&mutex); <===================== Add this
    }

    int main ()
    {
      int i;

      pthread_t tid;

      pthread_mutex_init(&mutex, NULL); <================= Add this

      for(i = 0; i < NTHREADS; i++)
      {

        pthread_create(&tid, NULL, thread, NULL);
      }

      pthread_exit(NULL);

      pthread_mutex_destroy(&mutex); <==================== Optional
    }



Question 8:


     Descriptor tables         Open file table       V-node table
     -----------------         ---------------       ------------

      parent:                     
        fd0          /----------> File pos
        fd1         /             refcnt = 1
        fd2-----\  /
        fd3----\ \/ 
        ...     \/\          
                /\ \    /-------->File pos
      child:   /  \ \  /          refcnt = 1
        fd0   /    \ \/
        fd   /      \/\
        fd2-/       /\_\=========>File pos
        fd3--------/              refcnt = 1
        ...



Question 9:

  Part 1:

    A. fd = socket(AF_INET, SOCK_STREAM, 0); 
       sin_port = htons(DEFAULT_PORT);
       s_addr   = htonl(INADDR_ANY);
       bind(fd, &addr, len);
       listen(fd, 10);
       return fd;

     B. fd = socket(AF_INET, SOCK_STREAM, 0);
        sin_port = htons(DEFAULT_PORT);
        s_addr   = 0x0100007F;  /* 127.0.0.1 in opposite endianness */
        connect(fd, &addr, len);
        return fd;


  Part 2:

       Client       Server
       -------------------
       1. socket
       2.           socket
       3.           bind
       4.           listen
       5.           accept
       6. connect


       There are many others, but it must follow these rules:

         * The server must call socket, bind, listen, accept in that order
         * Connect must happen after accept, otherwise connect could possibly
           timeout before being accept()ed.



Question 10:

  FD_SET (fd_list[j], &read_set);
  select (max_fd, &ready_set, NULL, NULL, NULL);
  FD_ISSET (fd_list[j], &ready_set);