Exam 1: Midterm

Solutions and Commentary

Part of the homework for 22C:60, Fall 2007
by Douglas W. Jones
THE UNIVERSITY OF IOWA Department of Computer Science

Grade Distributions

Exam Distribution

                                      X  
                                      X  
                                      X  
                                      X  
Mean   = 8.18                         X  
Median = 8.7                          X  
                                      X  
                                  X   X X
                          X       X X X X X
 ___________________X_____X___X_X_X_X_X_X_X_X_
   0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10

Homework Distribution, assignments 1 to 5

                                                        X
Mean   = 11.02                                    X   X X
Median = 11.5                   X           X X X X   X X
 _____________________X_____X___X___X_X___X_X_X_X_X_X_X_X_X_X_X___
   0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10. 11. 12. 13. 14. 15

Machine Problem Distribution

                        X
                        X
                      X X
                      X X
                      X X
Mean   = 4.63         X X
Median = 4.9          X X
                      X X
                      X X
    X                 X X
    X                 X X
 ___X_X_____________X_X_X_
   0 . 1 . 2 . 3 . 4 . 5

Overall Distribution

                                                  X
Mean   = 23.31                                    X       X
Median = 24.5                             X       X   X X X
                                      X   X       X X X X X   X
 _________________________X___X_______X___X___X___X_X_X_X_X___X___
   0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30
 Approximate midterm grade      F?     |    C/D    |    A/B    |

Exam Solutions

Fill in the following table; values in the blanks in each column derive from the given values in that column, all values in each row are derived from the given values using the same rules. (4 points)
```
Bit Pattern x         110000      001001      111010     011111     111011

Unsigned decimal     ___48___    ____9___    ___58___      31      ___59___

1's complement of x  _001111_    _110110_     000101    _100000_   _000100_ 

2's complement of x  _010000_     110111     _000110+   _100001_   _000101+

x as a 1's comp num    -15       ____9__*    ___-5__+   ___31__*   ___-4__+

x as a 2's comp num  __-16___    ____9__*    ___-6___   ___31__*      -5   
```
21 did this perfectly. 4 students each had difficulty with the answers marked with * above. Two student each had difficulty with the answers marked with + above. Some of these difficulties can be classified as clerical errors, but most of them appear related to misunderstandings of the signed number systems.

Translate the following brief useless SMAL code to a sequence of bytes loaded in memory. The spaces to the right are labeled with a word and byte in word addresses; fill in the hexadecimal value that goes in that byte; leave uninitialized bytes blank. The result of assembling the first lines are given. (2 points)

    . = 0                       ||  Address    3    2    1    0
        W     #01234567         ||          |----|----|----|----|
    ; --solved to here--        ||  000000: | 01 | 23 | 45 | 67 |
        B     A,B,C,D           ||          |----|----|----|----|
    A:  B     'A','B'           ||  000004: | 10 | 0E | 0C | 08 |
        ALIGN 4                 ||          |----|----|----|----|
    B:  B     #AB               ||  000008: |    |    | 42 | 41 |
        ALIGN 2                 ||          |----|----|----|----|
    C:  H     #0123             ||  00000C: | 01 | 23 |    | AB |
        ALIGN 4                 ||          |----|----|----|----|
    D:  B     #01,#23,#45,#67   ||  000010: | 67 | 45 | 23 | 01 |
        W     #01234567         ||          |----|----|----|----|
        END                     ||  000014: | 01 | 23 | 45 | 67 |
                                            |----|----|----|----|

14 students did perfectly here, 6 more made 5 or fewer errors. Word 4 of memory was the hardest because many students either interpreted the letters A, B, C and D as ASCII characters instead of labels, or they failed to correctly assign values to these labels. Problems with byte order and alignment were common.

a) Hand assemble the Hawk program fragment given below, assuming the first instruction goes in location 1000₁₆, giving the result as a sequence of halfwords in base 16 and using question marks for unknown values. The first instruction is done for you. (1.5 points)

                                    ||         001000:   F850
            LOAD    R8,PARRAY       ||
        ; --solved to here--        ||         001002:   ????
        LAB1:                       ||
            LOADSCC R3,R8           ||         001004: __F3C8__
            BZS     LAB2            ||
            ADDI    R8,R8,12        ||         001006: __0203__
            BR      LAB1            ||
        LAB2:                       ||         001008: __F868__
                                    ||
                                    ||         00100A: __000C__
                                    ||
                                    ||         00100C: __00FB__

6 did perfectly, about 10 had difficulties with branch displacements, and 7 failed to allocate an extra halfword for the operand of the ADDI instruction. Many made various clerical errors looking up opcodes.

b) What does this do> (The answer is very short.) (0.5 points)

_It_steps_through_an_array_of_3_word_records_until_it_finds_one_starting_with_0_

5 did well here, while 15 earned no credit at all. Explaining what a program does is difficult!

Complete the following SMAL Hawk program. When complete, it should have a function that conforms to the comments, and this function should be somewhat familiar! (2 points)

    ; record structure for an object
    TEXT    =       0
    X       =       4
    Y       =       8

    ; activation record structure for PUTOBJECT
    ; RETAD =       0
    P       =       4
    ARSIZE  =       8
    
    PUTOBJECT:
            ; on entry, R1 = return address
            ;           R3 = p, pointer to an object
            ; displays that object
            STORES  R1,R2           ; save return address
            STORE   R3,R2,P         ; save parameter p
            LOADCC__R0,R3,TEXT___   ; if (p->text != null)
            BZS     PUTDONE         ; {
            LOAD    R4,R3,Y         ;   parameter R4 = p->y
            LOAD____R3,R3,X______   ;   parameter R3 = p->x
            ADDI    R2,R2,ARSIZE
            LOAD____R1,PDSPAT____
            JSRS    R1,R1           ;   call dspat(p->x,p->y)
            ADDI    R2,R2,-ARSIZE
            LOAD    R3,R2,P
            LOAD____R3,R3,TEXT___   ;   parameter R3 = p->text
            ADDI    R2,R2,ARSIZE
            LOAD    R1,PDSPST
            JSRS____R1,R1________   ;   call dspst(p->text)
            ADDI    R2,R2,-ARSIZE
    PUTDONE:                        ; }
            LOADS   R1,R2           ; restore return address
            JUMPS   R1

3 did perfectly here, 20 earned 3/4 credit or more. The first blank posed the hardest problems. 15 used TESTR here, 7 more used CMPI -- in both cases, testing R3 instead of testing the memory location pointed to by R3. Note that there were good solutions submitted using LOADSCC and TESTS instruciton. It is difficult to classify the other errors students made, except to speculate that several were clerical.