Exam 1: Midterm

Commentary and Solutions

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

Grade Distributions

Exam

                                       X
                                       X
Mean   = 8.27                          X
Median = 8.5                           X
                                   X   X X
                                   X   X X
                                   X X X X     X
                       X           X X X X X   X
  _________________X___X_X___X___X_X_X_X_X_X_X_X_
    0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10.

Homeworks 1 to 5

Mean   = 10.75                               X         X
Median = 10.5                            X   X       X X     X
                                         X X X X   X X X     X
  ___X_______X___________X_______X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X___
    0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10. 11. 12. 13. 14. 15

Homeworks plus Exam

                                         X
Mean   = 18.70                       X X X     X
Median = 18.9                        X X X X X X X
                                     X X X X X X X X   X
  _____________X_________X_X_X_______X_X_X_X_X_X_X_X___X___
    0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20. 22. 24. 26
A very rough grade scale     F     D     C     B     A

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         101100      001101     010001     100111     010111

Unsigned decimal     ___44___    ___13___      17      ___39___   ___23___

1's complement of x  _010011_     110010    _101110_   _011000_   _101000_

2's complement of x   010100     _110011_   _101111_   _011001_   _101001_

x as a 1's comp num  __-19___*   ___13___   ___17___   __-24___*     23   

x as a 2's comp num  __-20___    ___13___*  ___17___*    -25      ___23___*

19 students did perfect work here, but 1/4 of the class made more than 3 errors. Stars above indicate problems where more than 6 students made mistakes.

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; use question marks for unknown or uninitialized bytes. 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 |
        A:  B     #01, #23    ||          |----|----|----|----|
            ALIGN 4           ||  000004: | ?  | ?  | 23 | 01 |
        B:  H     #ABCD       ||          |----|----|----|----|
            ALIGN 4           ||  000008: | ?  | ?  | AB | CD |
        C:  B     #34         ||          |----|----|----|----|
            ALIGN 2           ||  00000C: | 42 | 41 | ?  | 34 |
        D:  ASCII "ABCD"      ||          |----|----|----|----|
            ALIGN 4           ||  000010: | ?  | ?  | 44 | 43 |
            B     D,C,B,A     ||          |----|----|----|----|
            END               ||  000014: | 04 | 08 | 0C | 0E |
                                          |----|----|----|----|

Half the class scored above 1.6, 1/4 scored below 1.2. The hardest problem was getting the corrrect values for the labels A, B, C and D assembled into the 4 bytes at address 14₁₆; half the class had problems here. The next hardest problem was the ASCII text, wheree 1/4 of the class had difficulty. The ALIGN 2 directive also caused difficulty for a similar number.

Hand assemble the Hawk program fragment shown below, assuming the first instruction goes in location 100₁₆, giving the result as a sequence of halfwords in hexadecimal. The first instruction is done for you. (2 points)
```
                                    ||         000100:   D300
            LIS    R3,0             ||
        ; --solved to here--        ||         000102: __14CF__
        LAB1:                       ||
            ADDSI  R4,-1            ||         000104: __0102__ **
            BNS    LAB2             ||
            ADD    R3,R3,R5         ||         000106: __3335__
            BR     LAB1             ||
        LAB2:                       ||         000108: __00FC__ *
                                    ||
                                    ||         00010A: ________
```
Half the class earned at least 1.7, but 1/4 earned 11.4 or less. Stars indicate difficulty. Branch displacements were a problem for half the class, and getting the correct opcode for BNS was a problem for 1/4 of the class.

Extra Credit: What does this probgram fragment do (the answer is very short)? (1 point max)
```
__The program multiplies R4 times R5 and puts the result in R3__

__except when R4 is negative____________________________________
```
Few had perfect scores; the hardest part was noticing that this program does not multiply when R4 is negative. Most answers were long winded descriptions that said, in English, what the assembly code does, instruction by instruction, with no evidence of any decoding of the algorithm (the stupidest multiplication algorithm known to man).

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)

            SUBTITLE GCD - integer greatest common divisor function
    ; activation record structure
    RA      =       0
    ARSIZE  =       4
    ; -------------------
    GCD:    ; function called with return address in R1
            ; on entry
            ;   R3 unsigned positive integer i
            ;   R4 unsigned positive integer j
            ; on exit
            ;   R3 greatest common divisor of i and j

            STORES  R1,R2

            _CMP_____R3,R4______ * 
            BEQ     GCDQT           ; if (i != j) {
            BGT     GCDEL           ;   if (i < j) {
                                    ;     R4 = R4 - R3
            _SUB_____R4,R4,R3___

            _BR______GCDEI______
    GCDEL:                          ;   } else {
            SUB     R3,R3,R4        ;     R3 = R3 - R4
    GCDEI:                          ;   }
            _ADDSI___R2,ARSIZE__ *
            JSR     R1,GCD          ;   R3 = GCD(R3,R4)
            ADDSI   R2,-ARSIZE
    GCDQT:                          ; }
            _LOADS___R1,R2______
            JUMPS   R1              ; return R3

Half the class got 1.6 or better. 1/4 scored 1.0 or worse. Stars indicate particularly difficult instructions, but errors were scattered randomly over all instructions. Wrong addressing modes (indexed instead of short, for example) received partial credit.