CS 2733/2731, Org II, Spring 2004, First Exam, Answers
========================================================================
1.  (a) 46 (base 10) = 101110 (base 2)
       -46 (base 10 = ~101110 + 1 = 1111 1111 1101 0001 + 1 =
                                    1111 1111 1101 0010
    (b) 1011 1111 1101 1100  (48 more 0's)  (binary) =

        0 01111111101  1100  (48 more 0's)  (broken into fields) =

        (-1)^Sign x (1 + Significand) x 2^(Exponent - Bias) = 
        (-1)^1    x (1 + 0.750000000) x 2^(1021     - 1023) =   
        (-1)      x 1.75              x 2^(-2)              = 
           -(7/4)x(1/4) = -7/16 = -0.4375
========================================================================
2. 
    # CS 2733/2731, Computer Organization II, Spring 2004
    # MIPS program giving answer to Exam1, question 2
            .globl main
    main:	addu    $s7, $zero, $ra
    ########## Start of answer to Question 2 ######################
            .data                                                 #
    A:      .word 2, 3, 5, 7, 11, 13, 17, 19, 23, 0               #
            .text                                                 #
            la      $s0, A          # address of A                #
            addi    $t0, $0, 0      # loop counter, start at 0    #
            addi    $t1, $0, 10     # to terminate loop           #
            add     $t2, $0, $s0    # address of item in A        #
            addi    $s1, $0, 0      # running sum                 #
    Loop:   lw      $t3, 0($t2)     # store current $t0 into A    #
            addi    $t0, $t0, 1     # increment loop counter      #
            addi    $t2, $t2, 4     # increment pointer into A    #
            add     $s1, $s1, $t3   # add A[i] into running sum   #
            bne     $t0, $t1, Loop  # branch back to form loop    #
    ########## End of answer to Question 2 ########################
        ## Print the sum just calculated (100)
            li      $v0, 1
            add     $a0, $s1, $0
            syscall
            jal     Newl
    ########## Finish main###################
            addu    $ra, $zero, $s7
            jr      $ra
    ########## write newline #################################
    Newl:   li      $v0, 4
            la      $a0, Newline
            syscall
            jr      $ra
            .data	
    Newline:	.asciiz  "\n"
    ############### output #############################
    # % spim -file exam1.s
    # 100
    ####################################################
==========================================================================
3. 
# Answer to Exam1, Problem 3, Spring 2004 ####################
        .globl  main
main:   add     $s7, $0, $ra # save return address

########## First part of answer, call to Addup ##########
        addi    $a0, $0, 7
        addi    $a1, $0, 19
        jal     Addup
########## End of call to Addup #########################
        add     $t0, $0, $v0    # save result in $t0
        addi    $v0, $0, 1      # print the returned value
        add     $a0, $0, $t0
        syscall

        addi    $v0, $0, 4      # print a newline
        la      $a0, Newln
        syscall

        add     $ra, $zero, $s7 # restore return address
        jr      $ra
########## Second part of answer,  Code for Addup #######
Addup:
        addi    $sp, $sp, -4
        sw      $ra, 0($sp)
        add     $v0, $a0, $a1
        lw      $ra  0($sp)
        addi    $sp, $sp, 4
        jr      $ra
########## End of code for Addup ########################
        .data
Newln:  .asciiz "\n"
################# Output #######################################
four06% spim -file exam1_3.s
26    
===========================================================================
4.
# CS 2734, Computer Organization II, Spring 2004
       .globl main
main:
       addu   $s7, $zero, $ra

### First way uses addi and requires 0 for sign bit
       lui    $s3, 0x0344
       addi   $s3, $s3, 0x07ff
## output answer: 0x034407ff or 54790143
       li     $v0, 1
       move   $a0, $s3
       syscall
       jal    Newl
### Second way uses ori and and always works
       lui    $s3, 0x0344
       ori    $s3, $s3, 0x07ff
## output answer: 0x034407ff or 54790143
       li     $v0, 1
       move   $a0, $s3
       syscall
       jal    Newl
### Third method without lui:
       li     $s3, 0x0344
       sll    $s3, $s3, 16
       ori    $s3, $s3, 0x07ff
## output answer: 0x034407ff or 54790143
       li     $v0, 1
       move   $a0, $s3
       syscall
       jal    Newl
## final part of main
       addu    $ra, $0,  $s7
       jr      $ra
## Newl function, print a newline
Newl:  li     $v0, 4
       la     $a0, Newline
       syscall
       jr     $ra
       .data
Newline: .asciiz "\n"
#############################################
# Output:
# 54790143
# 54790143
# 54790143
   
==========================================================================
6. With the given inputs, the upper two transistors are open (don't conduct 
  current), while the bottom two are closed.  Thus C is connected to
  the ground at the bottom and not to any power, so C's output is 0.

   If either input is 0, C is 1, while both inputs 1 makes C a 0.
   Thus C is a NAND gate.