Exam 2

22C:18, Summer 1997, 12 points

Name ______________________________________   Section ___________

1. Fixed point numbers. In the following table, all numbers in each column share the same base, while all numbers in each row have the same value, fill in the blanks. (2.4 points)

       Base 2          Base 10    |    Base 2          Base 10          
                                  |
      1110.1011         14.6875   |
                                  |  _____._____         3.8125
      1101.0101      _____._____  |  
                                  |  _____._____         9.5625
      0111.1001      _____._____  |
   

2. Floating point numbers: Consider a system with the following format:

                            _ _ _ _ _ _ _ _
                           |_|_|_|_|_|_|_|_|
                           |s| exp |  man  |
   

   • s - the sign of the mantissa
   • exp - the exponent, a 3-bit 2's complement number.
   • man - the magnitude of the mantissa, fixed point, with a hidden bit.

   Answer the following questions for this number system, giving answers in conventional decimal form: (2.4 points)

   1. What is the largest possible exponent? ___________

   2. What is the smallest possible exponent? ___________

   3. What is the largest possible mantissa? ___________

   4. What is the smallest positive normalized mantissa? ___________

   5. What is the largest positive number in this system? ___________

   6. What is the smallest positive number in this system? ___________

3. Programming with various number representations: Write code to bridge the gap between the indicated pre and postconditions: (2.4 points)

        ; A is in R3, an 8 digit fixed point unsigned BCD number;
        ; A has 2 digits of fraction and 8 digits of integer part!
        ; B is pointed to by R8; B is a BCD integer.
        ; assume that BCDTOI and ITOBCD converts between 8 digit BCD
        ; numbers and binary integers in R3, and that they know nothing
        ; about fixed point numbers!
   
              CALL    BCDTOI    ; convert R3 to binary
   
              ____________________________
   
              ____________________________
   
              ____________________________
   
              CALL    ITOBCD    ; convert R3 back to decimal
   	     
        ; R3 now contains 10*A
   
              ____________________________
   
              ____________________________
   
              ____________________________
   
        ; R3 now contains 10*A rounded to the nearest integer
   
              ____________________________
   
              ____________________________
   
              ____________________________
   
        ; B now contains 10*A, in the format required for B
   

4. Given our standard activation record structure, where R2 points to the activation record that holds the current set of local variables, write code to bridge the gap between the following pre and post conditions: (2.4 points)

   	; A is a common holding a global array of integers;
   	; AP is the label on a word pointing to A[0].
   	;
   	; I is a common holding an integer;
   	; IP is the label on a word pointing to I.
   	;
   	; I is known to be within bounds for indexing into A.
   	;
   	; B is the displacement from R2 of a local array of
   	; integers.  B points to location 0 of that array.
   	;
   	; J is the displacement from R2 of a local integer.
   	;
   	; J is known to be within bounds for indexing into B.
   	
              ____________________________
   
              ____________________________
   	
              ____________________________
   
              ____________________________
   	
              ____________________________
   
              ____________________________
   
   	; R3 holds the value of A[I]
   
              ____________________________
   	
              ____________________________
   
              ____________________________
   	
              ____________________________
   
              ____________________________
   
   	; B[J] holds the value of A[I]
   

5. Procedures and functions! The following code is missing some key bits. Supply them! them (2.4 points):

        ; --------------------
        ; R2 points to AR with following format
        ;RA=    0  ; return address
        TIMEX=  4  ; extra value to add
        AR=     8  ; size of activation record
   
        TIMES:  ; horrible function to multiply two words
                ; algorithm:  A x B = if B zero 0
                ;                   = if B even B/2 x (2 x A)
                ;                   = if B odd (B/2 x (2 x A)) + A
                ; expects argument in R3, R4.
                ; returns result in R3.
                ; wipes out R5
   
   	     TEST    R4
   	     BZS     TIMQT     ; quit if B is zero
   
                _________________ ; save return address
   
   	     MOVE    R5,R3     ; \
   	     MOVE    R3,R4     ;  > swap A and B
   	     MOVE    R4,R5     ; /  (R5 ends up a copy of A)
   	     SRU     R4	       ; get B/2, set C if B odd
   	     BCS     TIMODD
   	     LIS     R5,0      ; if B even, R5=0 else R5=A
        TIMODD:
   	     _________________ ; set aside R5 as extra value
   
   	     SL	     R3,1      ; get 2xA
   
   	     _________________ ; \
   
   	     _________________ ;  > recursive call
   
   	     _________________ ; /
   
   	     _________________ ; Recover extra value in R4
   
   	     ADD     R3,R3,R4  ; add it to return value
   
                _________________ ; recover return address
   
        TIMQT:  JUMPS   R1