IC220 Computer Architecture
SOLUTION 6-Week Exam
Last Name ____________________ First Name _________________
Section:
(circle one)
1001
(Roth)
3001
(Rye)
3002
(McDowell)
5001
(Rye)
Alpha ___________________
5002
(McDowell)
Note: This exam is closed-book, closed-notes.
Exception: A reference sheet is attached to the last page. You may freely use this, but only
this sheet (not any other notes, or a copy of the sheet that you brought yourself).
No calculators are permitted.
To receive partial credit, show all work.
UNLESS STATED OTHERWISE, PSEUDO-INSTRUCTIONS MAY BE USED
Page 1 (15 pts)
______________
Page 2 (12 pts)
______________
Page 3 (10 9 pts)
______________
Page 4 (6 pts)
______________
Page 5 (11 pts)
______________
TOTAL(54 pts)
______________
NOTE: This is an exam that will be given to multiple sections and possibly to students after the
primary exam day. You may not discuss it with anyone until after Tues February 23, 2016.
SOLUTION IC220 6wk Exam – Mon Feb 22, 2016 SOLUTION
(1 pt) What is the tool that translates from a high-level language to assembly language?
Answer: _______COMPILER_____________________
(1 pts) Give an EXAMPLE of a particular “ISA” that is in use today.
Answer: ____x86, MIPS, SPARC, ARM________________________
(2.5 pts) List the five classic components associated with all computers. No description needed.
1. Input
2. Output
3. Memory
4. Control
5. Datapath
(2.5 pts) Suppose register $s4 holds the base address of array C (an array of 32-bit integers), and register $s2
holds the value of variable K (an integer). Show the MIPS instruction(s) for this C code:
C[4] = K + 10;
addi $t0, $s2, 10
sw $t0, 16($s4)
# Note – don’t modify $s2 – this code should NOT change the value of K!
(8 pts) For each of the following, circle whether it is a valid or invalid raw MIPS instruction (count pseudoinstructions as invalid). If invalid, briefly state why (must provide correct reason for credit if invalid).
lw $t0, $t1($s0)
VALID
INVALID because: lw/sw require register+offset
(not TWO registers for destination)
add $t0, $t1, 4($s2)
VALID
INVALID because: add requires 3 registers as operands
(can’t access memory with add)
addi $t0, $t0, -3
VALID
INVALID because:
blt $a0, $a1, Label17
(assume Label17 exists)
VALID
INVALID because: blt is pseudo-instruction
SOLUTION IC220 6 week Exam – Mon. Feb 22, 2016 SOLUTION
1
(6 pts) Consider this number (given in hex): 28A2000F
(3 pts) Show the BINARY form of this number:
0010 1000 1010 0010 0000 0000 0000 1111
Opcode is first 6 bits: 001010 = 00 1010 = 0A (hex). Which is “sltit” according to “green sheet.”
(3 pts) If that number represented a MIPS instruction, what would it be?
add
addi
and
beq
bne
j
jal
jr
lw
slt
slti
sw
sub
subu
(NO further details needed, just circle one)
(1 pts) (circle best answer)
In terms of the instructions available, which type of ISA tends to offer a larger number of complex,
powerful instructions? (circle one)
BISC
CISC
LISC
RISC
YISC
(5 pts) Below is a snippet of code. Assume that when it starts executing:
- $s0 holds the base address of some array of integers called “array1”
- $s1 holds the base address of some array of integers called “array2”
- $s3 holds an integer named “K”
What does this code do? (in terms of array1, array2, and K)
(we’re looking for a brief overall summary of the effect of executing code, NOT a description of how it
does it or what each instruction does):
L71:
move $a0, $s0
move $a1, $s1
add $t2, $zero, $zero
beq $t2, $s3, L73
lw $t0, 0($a0)
sw $t0, 0($a1)
addi $a0, $a0, 4
addi $a1, $a1, 4
addi $t2, $t2, 1
j L71
L73:
Your answer:
Copies first K elements (words) of array1 to array2.
SOLUTION IC220 6 week Exam – Mon. Feb 22, 2016 SOLUTION
2
(2.5 pts) On a certain machine, Program 1 runs in 10 seconds and requires 200 million cycles to execute.
 What is the clock rate of this machine? Make sure your answer has a unit.
Time = NumCycles / ClockRate
ClockRate = NumCycles / Time = 200 * 106 cycles / 10 seconds = 20 * 106 cycles = 20 MHz
(2.5 pts) Label the UNITS for each term in the equation below. Each term should have a unit for the numerator
and denominator (a total of FIVE units for you to fill in, using the blanks below the equation).
Time =



IC
*

inst

program

CPI



*
cyc
inst
ClockCycleTime

sec



cyc





(4 pts) Below are two examples of possible metrics for comparing performance. For each,
a. indicate whether using that metric, in the given context, is OKAY or BAD IDEA, *and*
b. briefly explain why.

The number of cycles needed to execute a program (comparing two versions of the same program on
the same machine)
a. Circle: OKAY or
BAD IDEA
b. Why?
Time = NumCycles * CycleTime
The cycle time (clock rate) is the same (same machine), so okay to just compare “NumCycles”

The number of instructions per second (comparing the same program on two different machines)
a. Circle: OKAY or BAD IDEA
b. Why?
The number of instructions needed (per program) may vary across the machines.
SOLUTION IC220 6 week Exam – Mon. Feb 22, 2016 SOLUTION
3
INSTRUCTIONS: If needed on this page and the next, assume the following variable/register mappings:
a – $s0
b – $s1
c – $s2
d – $s3
Where appropriate, be sure to obey register conventions.
It’s fine to use PSEUDO-INSTRUCTIONS (it always is unless we say explicitly not to)
(6 pts) Translate the following C++ code into MIPS assembly.
d = 0;
while (d < c) {
d = dragon(b);
b = b + 1;
}
Note – a “while” loop, so must “test first”
Loop:
li $s3, 0
j Test
move $a0, $s1
jal dragon
move $s3, $v0
addi $s1, $s1, 1
blt $s3, $s2, Loop
# or move $s3, $zero or similar
#
#
#
#
#
Or instead of blt:
slt $t0, $s3, $s2
bne $t0, $zero, Loop
prepare argument (b) for call
call dragon (b)
store return value into d
b++
keep looping if d < c
# is d < c
# if YES, continue looping
Alternate solution:
Loop:
li $s3, 0
bge $s3, $s2, Exit
move $a0, $s1
jal dragon
move $s3, $v0
addi $s1, $s1, 1
j Loop
#
#
#
#
#
#
or move $s3, $zero or similar
quit if d >= c (inverse of d < c)
prepare argument (b) for call
call dragon (b)
store return value into d
b++
Exit:
Or instead of bge above:
slt $t0, $s3, $s2
beq $t0, $zero, Exit
# is d < c?
# if NOT, exit
SOLUTION IC220 6 week Exam – Mon. Feb 22, 2016 SOLUTION
4
(11 pts) Translate the following C++ code into MIPS assembly. Follow all conventions.
int finn (int a, int b, int c) {
int t = 20;
return rey(c, t) + kylo(b, a) + c;
}
finn:
addi $sp, $sp, -20
sw $ra, 0($sp)
sw $a0, 4($sp)
sw $a1, 8($sp)
sw $a2,12($sp)
# call rey(c,t)
li $t0, 20
move $a0, $a2
move $a1, $t0
jal rey
sw $v0, 16($sp)
# call kylo(a,b)
lw $a0, 8($sp)
lw $a1, 4($sp)
jal kylo
# space for 5 items
# save a
# save b
# save c
# t0 = 20
# first argument = c
# second argument = t
# save rey’s result
# first argument = b
# first argument = a
# call kylo, result now in $v0
# add results together
lw $t0, 16($sp)
# get
lw $t1, 12($sp)
# get
add $v0, $v0, $t0
# add
add $v0, $v0, $t1
# add
rey’s result
c
rey’s result to kylo’s
c to overall result
# cleanup and return
lw $ra, 0($sp)
# reload return addres
addi $sp, $sp, 20
# fix up stack
jr $ra
# return
# NOTE – if you want to use $s0, $s1, etc. – must save original value
# and restore at the end!
SOLUTION IC220 6 week Exam – Mon. Feb 22, 2016 SOLUTION
5