Week 07 Tutorial Answers

  1. How is the assignment going?

    Does anyone have hints or advice for other students?

  2. Give MIPS directives to represent the following variables:

    1. int v0;
    2. int v1 = 42;
    3. char v2;
    4. char v3 = 'a';
    5. double v4;
    6. int v5[20];
    7. int v6[10][5];
    8. struct { int x; int y; } v7;
    9. struct { int x; int y; } v8[4];
    10. struct { int x; int y; } *v9[4];

    Assume that we are placing the variables in memory, at an appropriately aligned address, and with a label which is the same as the C variable name.

  3. Translate this C program to MIPS assembler.
    int max(int a[], int length) {
        int first_element = a[0];
        if (length == 1) {
            return first_element;
        } else {
            // find max value in rest of array
            int max_so_far = max(&a[1], length - 1);
            if (first_element > max_so_far) {
                max_so_far = first_element;
            }
            return max_so_far;
        }
    }
    
  4. Translate this C program to MIPS assembler using normal function calling conventions.

    sum2 is a very simple function but don't rely on this when implementing sum4.

    // sum 4 numbers using function calls
    
    #include <stdio.h>
    
    int sum4(int a, int b, int c, int d);
    int sum2(int x, int y);
    
    int main(void) {
        int z = sum4(11, 13, 17, 19);
        printf("%d\n", z);
        return 0;
    }
    
    int sum4(int a, int b, int c, int d) {
        int e = sum2(a, b);
        int f = sum2(c, d);
        return sum2(e, f);
    }
    
    int sum2(int x, int y) {
        return x + y;
    }
    
  5. For each of the following struct definitions, what are the likely offset values for each field, and the total size of the struct:

    1. struct _coord {
          double x;
          double y;
      };
      
    2. typedef struct _node Node;
      struct _node {
          int value;
          Node *next;
      };
      
    3. struct _enrolment {
          int stu_id;         // e.g. 5012345
          char course[9]:     // e.g. "COMP1521"
          char term[5];       // e.g. "17s2"
          char grade[3];      // e.g. "HD"
          double mark;        // e.g. 87.3
      };
      
    4. struct _queue {
          int nitems;     // # items currently in queue
          int head;       // index of oldest item added
          int tail;       // index of most recent item added
          int maxitems;   // size of array
          Item *items;    // malloc'd array of Items
      };
      

    Both the offsets and sizes should be in units of number of bytes.

Revision questions

The following questions are primarily intended for revision, either this week or later in session.
Your tutor may still choose to cover some of these questions, time permitting.

  1. Consider the following 3-d array structure:

    The cube could be implemented as an array of pointers, each of which points to a slice of the cube. Each slice of the cube is also an array of pointers, and each of those points to an array of int values.

    For example, in the diagram above, the cell labelled x can be accessed as cube[0][0][1], and the cell labelled y can be accessed as cube[0][2][3].

    1. Write MIPS assembler directives to define a data structure like this.
    2. Write MIPS assembler code to scan this cube, and count the number of elements which are zero.

    In other words, implement the following C code in MIPS.

    int cube[4][4][4];
    
    int nzeroes = 0;
    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 4; j++)
            for (int k = 0; k < 4; k++)
                if (cube[i][j][k] == 0)
                    nzeroes++;
    
  2. Challenge exercise, for those who have seen linked data structures in C.

    Consider a linked list

    which could be defined in MIPS as:

        .data
    list:   .word node1
    node1:  .word 6, node2
    node2:  .word 4, node3
    node3:  .word 5, node4
    node4:  .word 2, 0
    

    Write a MIPS function that takes a pointer to the first node in the list as its argument, and returns the maximum of all of the values in the list. In other words, write a MIPS version of this C function:

    typedef struct _node Node;
    struct _node { int value; Node *next; };
    
    int max (Node *list)
    {
        if (list == NULL) return -1;
        int max = list->value;
        Node *curr = list;
        while (curr != NULL) {
            if (curr->value > max)
                max = curr->value;
            curr = curr->next;
        }
        return max;
    }
    

    You can assume that only positive data values are stored in the list.

  3. If we execute the following small MIPS program:

       .data
    x: .space 4
       .text
       .globl main
    main:
       li  $a0, 32768
       li  $v0, 1
       syscall
       sw  $a0, x
       lh  $a0, x
       li  $v0, 1
       syscall
       jr  $ra
    

    ... we observed that the first syscall displays 32768, but the second syscall displays -32768. Why does this happen?

  4. FIFO queues can be implemented using circular arrays. For example:

    And the C code to manipulate such a structure could be:

    // place item at the tail of the queue
    // if queue is full, returns -1; otherwise returns 0
    int enqueue (int item)
    {
        if (nitems == 8) return -1;
        if (nitems  > 0) tail = (tail + 1) % 8;
        queue[tail] = item;
        nitems++;
        return 0;
    }
    
    // remove item from head of queue
    // if queue is empty, returns -1; otherwise returns removed value
    int dequeue (void)
    {
        if (nitems == 0) return -1;
        int res = queue[head];
        if (nitems  > 1) head = (head + 1) % 8;
        nitems--;
        return res;
    }
    

    Assuming that the items in the queue are ints, and the following MIPS data definitions are used:

    nitems: .word 0
    head:   .word 0
    tail:   .word 0
    items:  .space 32
    

    ... implement the enqueue() and dequeue() functions in MIPS.

    Use one of the standard function prologues and epilogues from lectures.