Exam 1: Midterm

Solutions and Commentary

Part of the homework for 22C:60, Fall 2005
by Douglas W. Jones
THE UNIVERSITY OF IOWA Department of Computer Science

Grade Distributions

Midterm 1

                                           X
Mean   = 7.57                              X
Median = 7.9                               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 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10

Machine Problems 1 and 2

                                               X
                                               X
Mean   = 7.32                                  X
Median = 8.0                                   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

Homeworks 1 through 5

Mean   = 10.6                                          X
Median = 11.3                                          X 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 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 .10 .11 .12 .13 .14 .15

Overall

Mean   = 24.80                               X
Median = 27.3                                X       X   X   X
                             X               X       X   X X X
_________X_______________X___X_X_X_X_________X___X___X_X_X_X_X___X_X___
  2 . 4 . 6 . 8 .10 .12 .14 .16 .18 .20 .22 .24 .26 .28 .30 .32 .34 .36
                       F         D         C         B         A

Exam

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         011101      011000     101010     111001     100101

Unsigned decimal     ___29___    ___24___      42      ___57___   ___37___

1's complement of x  _100010_     100111    _010101_   _000110_   _011010_

2's complement of x   100011     _101000_   _010110_   _000111_   _011011_

x as a 1's comp num  ___29___    ___24___   __-21___   ___-6___     -26   

x as a 2's comp num  ___29___    ___24___   __-22___      -7      __-27___
```
14 did perfect work here. 5 gave strange values for all of the positive 2's complement numbers, despite warnings in class about this problem. Overall, the 1's and 2's complement number systems posed the greatest challenge. 3 students had trouble with carry propagation in performing the 2's complement operation, and a few students made clerical errors elsewhere.

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 |
            H     #5678       ||          |----|----|----|----|
            ALIGN 4           ||  000004: |    |    | 56 | 78 |
        A:  B     #89,#67     ||          |----|----|----|----|
            ALIGN 4           ||  000008: |    |    | 67 | 89 |
        B:  B     #9A         ||          |----|----|----|----|
            ALIGN 2           ||  00000C: | 31 | 30 |    | 9A |
        C:  ASCII "0123"      ||          |----|----|----|----|
            ALIGN 4           ||  000010: |    |    | 33 | 32 |
        D:  B     A,B,C,D     ||          |----|----|----|----|
            END               ||  000014: | 14 | 0E | 0C | 08 |
                                          |----|----|----|----|

8 did perfect work here. 9 had serious problems with the string "0123", and 9 had serious problems with the label values A, B, C and D. Because of the ALIGN 2 directive, the label C was particularly difficult, giving problems for another 6.

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. The first instruction is done for you. (1.5 points)

                                    ||         001000:   D300
            LIS    R3,0             ||
        ; --solved to here--        ||         001002: __F0E4__
        LAB1:                       ||
            TESTR  R4               ||         001004: __0203__
            BZS    LAB2             ||
            ADDSI  R3,1             ||         001006: __13C1__
            LOADS  R4,R4            ||
            BR     LAB1             ||         001008: __F4D4__
        LAB2:                       ||
                                    ||         00100A: __00FB__

3 did perfect work here. 12 had difficulty with the first branch displacement (the forward branch), while 15 had difficulty with the second one (the backward branch). Aside from this, there were a scattering of clerical errors.

b) What value does this put in R3 (the answer is very short)? (0.5 points)

__ the length of the chain of pointers referenced by R4 _______________________

3 did perfect work here. 18 earned no credit. Partial credit was given for answers such as "the number of iterations of the loop"

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
    NEXT    =       0
    X       =       4
    Y       =       8
    TEXT    =       12

    ; activation record structure for PUTOBJECT
    RETADDR =       0
    P       =       4
    ARSIZE  =       8
    
    PUTOBJECT:
            ; on entry, R1 = return address
            ;           R3 = p, pointer to an object
            ; displays that object, unless p = null
            STORES  R1,R2           ; save return address
            STORE   R3,R2,P         ; save parameter p
            TESTR   R3

            _BZS_____PUTDONE_______ ; if (p != NULL) {
            LOAD    R4,R3,Y         ;    parameter p->y
            LOAD    R3,R3,X         ;    parameter 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 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____________ ; return

4 did perfect work here. 7 had no clue about the ADDI -ARSIZE instruction. 8 had trouble with register usage on the LOAD TEXT instruction, and 8 had trouble with the return, mostly using JSRS instructions to do it. Other errors were more difficult to classify.