Exam 1: Midterm

Score Distribution

Median = 5.8
                   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
Rough Scores: D       C       B       A

1. 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         101010      111010     110010     010111     010011
   
   Unsigned decimal     ___42___    ___58___      50      ___23___   ___19___
   
   1's complement of x  _010101_     000101    _001101_   _101000_   _101100_
   
   2's complement of x   010110     _000110_   _001110_   _101001_   _101101_
   
   x as a 1's comp num  __-21___    ___-5___   __-13___   ___23___      19  
   
   x as a 2's comp num  __-22___    ___-6___   __-14___      23      ___19___
   

   5 did perfectly here, an equal number had extreme problems with the interpretation of x as a one's or two's complement number, getting all the answers in the final two rows wrong.

2. 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 bytes blank that do not receive values. 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     A, B, C, D  ||          |----|----|----|----|
               ALIGN 2           ||  000004: | 10 | 0C | 08 | 04 |
           B:  W     0           ||          |----|----|----|----|
           .   =   B             ||  000008: | 00 | 00 | FF | FF |
               H     -1          ||          |----|----|----|----|
               ALIGN 4           ||  00000C: | 33 | 32 | 31 | 30 |
           C:  ASCII "0123"      ||          |----|----|----|----|
               ALIGN 4           ||  000010: |    |    | 35 | 34 |
           D:  B     "4", "5"    ||          |----|----|----|----|
               END               ||  000014: |    |    |    |    |
                                             |----|----|----|----|
   

   At least 6 earned no credit here. Common errors included failing to notice that A, B, C and D are labels, failing to notice the effect of the .=B line, failing to correctly interpret the ALIGN directives, and failing to properly convert ASCII to hex.

3. 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:   E301
               LIL    R3,#1FFFC        ||         000102:   FFFC
           ; --solved to here--        ||
               LIL    R4,#0FFFF        ||         000104: __E400__
           LP:                         ||
               STORES R3,R3            ||         000106: __FFFF__
               ADDSI  R3,-4            ||
               CMP    R3,R4            ||         000108: __F3A3__
               BGT    LP               ||
                                       ||         00010A: __13CC__
                                       ||
                                       ||         00010C: __2034__
                                       ||
                                       ||         00011E: __0EFC__
   

   Extra Credit: What does this probgram fragment do (the answer is very short)? (1 point max)

   __it sets each word in RAM to hold its address _________________
   

   At least 4 did perfect work here. One earned no credit at all, and the remainder made mistakes that varied from the careless to the odd.

4. Complete the following SMAL Hawk program, conforming to the comments. (2 points)

               SUBTITLE TIMES - multiply
       RA      =       0	; activation record structure
       I       =       4
       J       =       8
       ARSIZE  =       12
       ; -------------------
       TIMES:  ; on entry, R3 and R4 are i and j, unsigned integers
               ; on exit R3 is i * j
               STORES  R1,R2
   
               _STORE  R3,R2,I_________
   
               _STORE  R4,R2,J_________
               STORE   R4,R2,J
               TESTR   R4
               BZR     MULDO           ; if (j == 0) {
   
               _LIS    R3,0____________
               JUMPS   R1              ;   return 0
       MULDO:                          ; }
               SL      R3,1
               
               _SR     R4,1____________
               ADDI    R2,R2,ARSIZE
   
               _JSR    R1,TIMES________
               ADDI    R2,R2,-ARSIZE   ; R3 = times( i << 1, j >> 1 )
   
               LOADCC  R4,R2,J
               TBIT    R4,0
               BCR     MULQT           ; if (odd(j)) {
   
               _LOAD   R4,R2,I_________
               ADD     R3,R3,R4        ;   R3 = R3 + i
       MULQT:                          ; }
               _LOADS  R1,R2___________
                                       ; return R3
               _JUMPS  R1______________
   

   Only one student earned no credit here, many earned half credit and a few earned 3/4 credit. Careless errors were common, but some student seemed intent on using instructions from some other algorithm, as if this might have been code from the book. It was not.