(10 pts) Carefully do assigned reading for Chapter 1 (1.1-1.5, 1.7-1.8)
(especially important b/c much of this chapter not covered in class).
Then answer the following questions.
 Exercise 1.1 from the book (see end of chapter). Place your
answers below (list and briefly describe).
1.
2.
3.
4.
(3 pts) Exercise 1-2
•
Do exercise 1.2 from the textbook. Follow example below, where some
answers are provided for you. Use each term exactly once.
a.
b.
(Hint: think about what these cables are made of)
c. Performance via Prediction
d.
e. Hierarchy of Memories
(2.66 x 109) / (12.5 x 106) = 212.8
f.
95W / 3.3 W = 28.8
g.
h.
(5 pts) Exercise 2-1
•
What is the MIPS assembly code for the following:
g = g + h - i;
Variables g, h, & i are assigned registers $s1, $s2, and $s4
add $s1, $s1, $s2 # g = g + h
sub $s1, $s1, $s4 # g = g - i
(5 pts) Exercise 2-2
•
What is the MIPS assembly code for the following:
g = h + A[3];
Variables g, h, & i are assigned registers $s1, $s2, and $s4
Array A base address is assigned register $s3
lw $t0, 12($s3)
add $s1, $s2, $t0
# temp = A[3]
# g = h + A[3]
(Note
(5 pts) Exercise
2-3different index calculation here)
•
What is the MIPS assembly code for the following:
g = h + A[i];
Variables g, h, & i are assigned registers $s1, $s2, and $s4
Array A base address is assigned register $s3
What is address of A[i]? (4*i + s3)
add $t1, $s4, $s4
# temp = 2 * I
add $t1, $t1, $t1
# temp = 4 * I
add $t1, $t1, $s3
# temp = &A[i]
lw $t0, 0($t1)
# temp = A[i]
add $s1, $s2, $t0
# g = h + A[i]
(extra space)
(10 pts) Exercise 2-4: Assume variables a, b, and c are assigned registers $s1, $s2,
and $s3, and the address of array A is in $s5. Write the code for the following:
b = A[4] - A[5];
lw $t0, 16($s5)
lw $t1, 20($s5)
sub $s2, $t0, $t1
# $t0 = A[4]
# $t1 = A[5]
# b = A[4] – A[5]
(10 pts) Exercise 2-5: Assume variables a, b, and c are assigned registers $s1, $s2,
and $s3, and the address of array A is in $s5. Write the code for the following:
b = A[2 * c];
Address of A[2 * c] = (Base of A)+2*4*c = $s6 + 8*c
add $t0, $s3, $s3
# $t0 = 2*c
add $t0, $t0, $t0
# $t0 = $t0 + $t0 = 4*c
add $t0, $t0, $t0
# $t0 = $t0 + $t0 = 8*c
add $t0, $t0, $s5
# $t0 = $s6 + 8*c
lw $s2, 0($t0)
# $t0 = Mem[$s6+8c] = A[2*c]
Or use sll to do multiplication. Shift by 3 bits to multiply
by 8:
sll $t0, $s3, 3
# $t0 = 8*c
add $t0, $t0, $s5
# $t0 = $s6 + 8*c
lw $s2, 0($t0)
# $t0 = Mem[$s6+8c] = A[2*c]
(6 pts) Exercise 2-8
(See number discussion in Section 2.5)
• What binary number does this hexadecimal number represent:
7fff fffahex?
•
What decimal number does it represent?
Easiest way to compute is to use hex:
= 7 * 167 + 15 * 166 + …. + 15 * 161 + 10 * 160
(10 pts) Exercise 2-9
Show the hexadecimal representation of this MIPS instruction:
add
$t0 ,
$t1 ,
$zero
Consider the format of the R-type instruction
op
6 bits
rs
5 bits
rt
5 bits
rd
shamt funct
5 bits
5 bits
6 bits
Here we have:
opcode = 0
rs = 9
rt = 0
funct = 0x20
rd = 8
shamt = 0
The bit string is: 000000 01001 00000 01000 00000
Rewriting, grouping bits four at a time:
0000 0001 0010 0000 0100 0000 0010 0000
In hex: 0x01204020
100000
(10 pts) Exercise 2-10
What MIPS instruction is represented by this binary entry:
(Tip: start by figuring out what the opcode is, then the instruction type)
1000 1101 0000 1001 0000 0000 0100 0100
Start by looking at the opcode (the first six bits): 100011. This is
equivalent to 0x23.
This is the opcode for load word: lw, which is an I-type instruction, whose
format is:
op
6 bits
rs
5 bits
rt
constant or address
5 bits
16 bits
Looking at the bit values, we see that rs = 8, rt = 9 and the constant is 68.
From the table, we see reg #8 is $t0, and reg #9 is $t1. Thus:
lw
$t1 , 68($t0)
(5 pts) Exercise 2-11
•
What is the MIPS assembly code for the following:
if (g != j) h = g - h;
else
h = g + h;
Variables f to j are assigned to registers $s0 to $s4
beq
sub
j
Else: add
Exit:
$s1, $s4, Else
$s2, $s1, $s2
Exit
$s2, $s1, $s2
….
# go to Exit
f
g
h
i
j
$s0
$s1
$s2
$s3
$s4
(5 pts) Exercise 2-12
•
What is the MIPS assembly code for the following:
if (j == h) g = i + j;
Variables f to j are assigned to registers $s0 to $s4
f
g
h
i
j
$s0
$s1
$s2
$s3
$s4
bne $s4, $s2, Exit
add $s1, $s3, $s4
Exit:
….
(5 pts) Exercise 2-13
•
Notice more advanced control flow.
What is the MIPS assembly code for the following:
if ( (j == h) && (f != i) ) g = i + j;
Variables f to j are assigned to registers $s0 to $s4
bne $s4, $s2, Exit
beq $s0, $s3, Exit
add $s1, $s3, $s4
Exit:
….
f
g
h
i
j
$s0
$s1
$s2
$s3
$s4
(10 pts) Exercise 2-14
•
What is the MIPS assembly code for the following:
if ( ( (g != h) && (f == i) ) ||
(g == i) )
g = i + j;
Variables f to j are assigned to registers $s0 to $s4
f
g
h
i
j
$s0
$s1
$s2
$s3
$s4
Many solutions possible, here is one
beq $s1, $s2, SecondChance
# check g != h
bne $s0, $s3, SecondChance
# check f == i
j DoAdd
SecondChance: # Didn’t pass first test, try second
bne $s1, $s3, Exit
# check g == i
DoAdd:
add $s1, $s3, $s4
# really do the add
Exit:
….
Here is another solution:
beq $s1, $s3, DoAdd
beq $s1, $s2, Exit
bne $s0, $s3, Exit
DoAdd:
add $s1, $s3, $s4
Exit:
….
(blank)
# check g == i
# check g != h
# check f == I
# really do the add