CS/COE 447 Quiz 5  February 26, 2007
====================================

Q1:   Assume we are using 8 bits to represent integers.

Consider the following operation:

     11001000
     01000000
   + --------
     00001000

Q1.A (4 points): 

Treating these values as unsigned integers, does unsigned overflow
occur as a result of this operation?  How do you know? Please explain
your answer.

Yes.  This is unsigned overflow:  viewing the numbers as unsigned
integers, the sum is too large to fit into 8 bits.  The maximum
unsigned value that can fit into 8 bits is (2^8) - 1, which is 255.
The sum, on the other hand, is 264 (200 + 64).  [Specific numbers were
not needed in your answers]

We can detect this overflow in different ways:

  (1) there is a carry out of the top bit
  (2) the sum is larger than the first operand   (a homework 2 question)
  (3) the sum is larger than the second operand  (a homework 2 question)


Q1.B (4 points):  

Treating the values as signed integers, does SIGNED overflow
result from the operation?  Please explain your answer.

No, signed overflow does not occur.  The first operand is a negative
number and the second is a positive number.  Signed overflow cannot
occur as the result of adding a positive and a negative number.  

=================================================================

Q2 (4 points): When the following code is executed, what hex values
are placed into $t1,$t2, $s1, and $s2?

.data
nums:
.word 0x00008888,0x88888888
.text
la $t0,nums
lw $t1,0($t0)
lw $t2,4($t0)
slt $s1,$t1,$t2
sltu $s2,$t1,$t2

$t1 = 0x00008888

$t2 = 0x88888888

$s1 = 0, since the first is a positive number and the second a
      negative number, so the first is not smaller than the second

$s2 = 1, since the first is a smaller unsigned number than the second
      one is


=================================================================
Q3 (8 points):  Assume the datapath is only 4 bits.  We will multiply 
two 4-bit unsigned integers, specifically:

    0100   
  X 0010
  ------
00001000

Using the algorithm on slides 15 and 16 of the lecture notes for
Chapter 3, Part 2 (the one we announced would be on this quiz),
perform the operation.  Specifically, complete the table below:

Iteration                  Multiplicand              Product
---------                  -------------             -------

   0 - Initial values:      0100                  0000 0010  

   ---------------------------------------------------------------
   1                        0100                  0000 0001 

   ---------------------------------------------------------------
   2                        0100                  0100 0001 
                                                  0010 0000
   ---------------------------------------------------------------
   3                        0100                  0001 0000 
   
   ---------------------------------------------------------------
   4                        0100                  0000 1000

   ---------------------------------------------------------------