COE 538
Quiz
Notes:
1.
Close book
2.
Write the answers in the space provided
3.
Show the process that is used to derive your answers
1.
Write two assembler directives to initialize memory at
locations $6000 and $6001 to $1025.
[1 mark]
ORG $6000
FDB $1025
2.
Given bit N of the Condition Code Register (CCR) is
set to 1, bit V is set to 1, and the Program Counter (PC) is set to
$6000. Calculate the value stored in PC after the following instruction is
executed:
bge $20 [2
marks]
PC = $6020
3.
Given the value stored in Accumulator A is $18 and the value
stored in Accumulator B is %01001100, calculate the value stored in Accumulator
D. Write down the value using decimal representation. [2
marks]
0001 1000 0100 1100 2 = 6220
10
4.
Write one instruction to accomplish each of the following
tasks: [4
marks]
a)
Add the content of Accumulator B to Accumulator A.
b)
Branch to memory location [PC] + $60.
BRA $60
c)
Set the most significant three bits of Accumulator B to 1.
ORAB #%1110 0000
d)
Clear the least significant two bits of Accumulator A to 0.
ANDA #%111 1100
5.
Given the program listing below, trace the results for each
instruction from start to end. Use the table provided below to
indicate the values stored in register A, B, D, X, and memory location $6050
and $6051 after the execution of each instruction. Show all numbers using
hexdecimal representation. [6 marks]
org $6000
values fcb $23, $D1, $A2, $3F
org $6050
result fcb $00, $00
org $6100
start clra a=0
clrb b=0
ldx #values x=$6000
ldd values d=$23D1
addb 1,x b=A2, c=1
adca 2,x a=C6=1100 0110
asra a=1110 0011=E3
std result [6050]=$E3 [6051]=$A2
end swi
|
Instructions |
A |
B |
D |
X |
[$6050] |
[$6051] |
|
ldx
#values |
00 |
00 |
00 |
6000 |
00 |
00 |
|
ldd
values |
23 |
D1 |
23D1 |
6000 |
00 |
00 |
|
addb
1,x |
23 |
A2 |
23A2 |
6000 |
00 |
00 |
|
adca
2,x |
C6 |
A2 |
C6A2 |
6000 |
00 |
00 |
|
asra |
E3 |
A2 |
E3A2 |
6000 |
00 |
00 |
|
std
result |
E3 |
A2 |
E3A2 |
6000 |
E3 |
A2 |
6.
Write an instruction sequence to find the sum of the first N
numbers in the following number sequence: 2, 4, 6, 8, 10… [10
marks]
CLRA
CLRB
m1 ADDA #2
INCB
CMPB #N
BNE m1
SWI