CS 354, version A
Fall 2004 		ANSWER KEY
Exam 1

No electronic devices may be used while taking this exam. Examples of devices not allowed are calculators, pagers, cell phones, wrist calculators/computers, laptop computers, pocket computers. Each student is allowed one 8.5 by 11 inch sheet of paper with handwritten notes. The notes may be on both sides of the paper.

Show all work, and do any/all calculations on the exam. Extra scratch paper may not be used.

Exam Score
Q1 = _____ /    8
Q2 = _____ / 10
Q3 = _____ /    8
Q4 = _____ / 10
Q5 = _____ / 10
Q6 = _____ / 10
Total = _____ / 56		 




Question 1 (8 points)



Part A
Assume that we have a 1-bit unsigned integer representation.
List each possible representation, and give the 
decimal value or values represented for each bit pattern.

   representation    decimal integer value
   --------------    ---------------------

          0                    0
	  1                    1






Part B
Same question as in Part A, but with a 1-bit two's complement
representation.

   representation    decimal integer value
   --------------    ---------------------


          0                    0
          1                    -1







Question 2 (10 points)



Part A
Identify the six steps of the instruction fetch and execute cycle
in their proper order:

    

  1. )
instruction fetch 

     
    

  2. )
PC update

     
    

  3. )
decode instruction

     
    

  4. )
load operands

     
    

  5. )
do the operation specified by the instruction

     
    

  6. )
store result(s)





Part B
Which of these steps could be placed at a different position
within the six steps?



Of the 6 steps, only step 2, PC update, can be placed at a
different position.
It could be placed at any position within the 6.





Question 3 (8 points)


A 354 student has written this SAL code fragment:



          .data
loop_count:   .word   0
sum:          .word   0
end_value:    .word

          .text

          move sum, 0
for_loop: move loop_count, 1
          bgt  loop_count, end_value, end_for
	  add  sum, sum, loop_count
	  add  loop_count, loop_count, 1
	  b    for_loop
end_for:





This code fragment is supposed to
sum all integers from 1 up to and including the value in the
variable
end_value.
Assume that the value of variable end_value is
set to a positive integer value in an earlier part of the program
that is not within this code fragment.


 

Part A
The code does not work.  What behaviour will this code fragment
exhibit when executed?



The code remains in an infinite loop, since the initialization
of the loop induction variable (loop_count) is
within the loop.
The code will likely cause an overflow exception when the value
of sum overflows.



Part B
The incorrect code fragment is repeated here.
Changing as few things as possible, show how to modify the code
fragment to make it work correctly.




          .data
loop_count:   .word   0
sum:          .word   0
end_value:    .word

          .text

          move sum, 0

          move loop_count, 1    # label removed from this instruction

for_loop: bgt  loop_count, end_value, end_for

	  add  sum, sum, loop_count

	  add  loop_count, loop_count, 1

	  b    for_loop

end_for:






 



Question 4 (10 points)


Part A
Give the decimal value 20.3 in base 4.
Show any repeating digits by placing a bar over those digits that repeat.


      Q  R
20/4  5  0
5/4   1  1
1/4   0  1
So, 20 (base 10) is 110 (base 4)


.3 x 4   1.2      1
.2 x 4   0.8      0
.8 x 4   3.2      3
.2 x 4 (repeats)
                      __
So, .3 (base 10) is .103 (base 4); only the last two digits repeat

Full answer:
                       __
20.3 (base 10) is 110.103 (base 4)





Part B
Give the octal value 60.2 in decimal.
Show any repeating digits by placing a bar over those digits that repeat.


6 x 8^1  +  0 x 8^0  +  2 x 8^(-1)
  48     +    0      +    .25

            48.25





Question 5 (10 points)


Give the  IEEE single precision floating point representation 
(in hexadecimal) for the decimal value

   -24.25.



24 (decimal) is 11000 (binary)

.25 (decimal) is .01 (binary)

11000.01

Move the radix point to put this in normalized form.

1.100001  X 2^4
  |
  ----> the mantissa portion of the answer

The exponent is 4. 
4 + 127 =  131

131 in 8-bit, unsigned representation is

10000011

In binary, putting the pieces together:

  S   E          F
  1  10000011   10000100000000000000000

Respacing, and then translating to hexadecimal:
  1100 0001 1100 0010 0000 0000 0000 0000

0x  c    1    c    2    0    0    0    0






Question 6 (10 points)


Part A

Write a SAL code fragment along with all variable declarations
necessary
for a SAL procedure named incr_aa that increments
an integer variable
named aa.




   .data
incr_aa_ra:  .word
aa:          .word

   .text
incr_aa:     add  aa, aa, 1
             b    (incr_aa_ra)





Part B
Write a SAL code fragment that uses incr_aa
to increment a variable named xx.




   .data
xx:         .word

   .text
            move aa, xx
            la   incr_aa_rtn, rtn
            b    incr_aa
rtn:        move xx, aa