Download Problem 1 - WSU EECS

EE-314
Spring 2001
Midterm Exam #1 Answer Key
Name:_____________________________________
Student ID #:_______________________________
Problem 1) (5 Points)
Identify the operand addressing mode used in each of these instructions.
a)
AND
DX,CX
__Register____________
b)
CALL
WORD PTR [SI+4]
__Index + Disp________
c)
ADD
CX,15
__Immediate__________
d)
MOV
IVAL[BX+DI+2],BX
__Base + Index +Disp___
e)
CMP
FOO[BP],AL
__Base + Disp_________
Problem 2) (10 Points)
What will be the value in AX after executing the following instructions? Give the answer
in both hexadecimal and binary.
mov
xor
mov
rol
add
adc
al,17
ah,ah
cl,4
al,cl
al,88h
ah,0
AX
AX
AX
AX
AX
AX
=
=
=
=
=
=
??
00
00
00
00
00
11
11
11
11 CY = 1
99 CY = 0
99
__0099h_______________
__0000 0000 1001 1001b__
Problem 3) (10 Points)
What will be the value in AX after executing the following instructions? Assume that DS
and ES are set up appropriately to access the variable ‘stuff’. Give the answer in
hexadecimal:
1
stuff
0
3
2
5
4
7
6
dw
12 34h,23 45h,34 56h,45 67h
mov
mov
add
mov
bx,1
si,5
si,bx
ax,stuff[si-3]
BX=1
SI=5
SI=6
AX=[stuff+3] (byte 3 & 4)
______5623h___________
AX gets the contents of bytes at offset 3 and 4 from the beginning of stuff. The byte at
offset 3 goes to AL and the byte at offset 4 goes to AH.
EE-314
Spring 2001
Problem 4) (20 points) Write a complete DOS assembly language program which does
the following: Read a single character from the keyboard. If the character read is ‘A’,
print the message “A was entered”, if the character read was ‘B’, print the message “B
was entered”. If any other character was typed, print the message “Unknown character”.
The program should then terminate.
Hints: Make sure that the data is in a data segment and the code in a code segment. Set up
segment registers appropriately. Use appropriate DOS functions for the input and output.
_data
segment
msgA
msgB
msgErr
db
db
_data
ends
_stck
segment
stck
dw
128 dup (?)
label word
_stck
ends
_text
segment
assume
word public ‘code’
cs:_text,ds:_data,ss:_stck
main:
mov
mov
mov
ax,seg _DATA
ds,ax
es,ax
mov
int
ah,1
21h
cmp
jnz
mov
jmp
al,’A’
;was A typed
main10
dx,offset msgA
main50
main10:
cmp
jnz
mov
jmp
al,’B’
;was B typed
main20
dx,offset msgB
main50
main20:
mov
dx,offset msgErr
main50:
mov
int
ah,9
21h
mov
int
ax,4C00h
21h
_TEXT
byte public ‘data’
“A was entered”,0Dh,0Ah,’$’
“B was entered”,0Dh,0Ah,’$’
db
“Unknown character”,0Dh,0Ah,’$’
word stack ‘stack’
ends
end
main
;read a char from keyboard
;print the message
EE-314
Spring 2001
Problem 5) (20 Points) Write a subroutine called ‘SUM’ which accepts two word sized
parameters on the stack, adds them together and returns the result in AX. It should be a
near procedure declared using the PROC directive. (Hint: be sure to set up a stack frame
appropriately for accessing the parameters)
A call to this subroutine would appear as follows:
mov ax,val1
push ax
;push first parameter
mov ax,val2
push ax
;push second parameter
call
sum
;returns result in AX
add
sp,4
;clean up stack
sum
sum
proc
near
push
mov
bp
bp,sp
mov
add
ax,[bp+6]
ax,[bp+4]
pop
ret
bp
endp
;get val1
;add in val2
EE-314
Spring 2001
Problem 6) Refer to the information on the following page to answer these questions:
a) (10 Points) The instruction shown in bold (address: 1266:0043) in the program
listing is the current instruction being executed. While this instruction is
executing, an NMI occurs. The NMI will be serviced before the next instruction
begins executing. What is the address of the NMI interrupt service routine?
NMI uses the INT 2 vector. Each vector occupies four bytes, so the INT 2
vector is the four bytes at offset 8 in the interrupt vector table (0000:0008).
Remember the proper byte order for Intel machines.
___05DB:1016___________
b) (15 Points) Show the contents of the program stack in the NMI interrupt service
routine just before the call to HANDLENMI occurs. Use one row of the table for
each word on the stack. (Hint: There will be values placed on the stack by the
CPU during interrupt processing and by the ISR itself before the call to
HANDLENMI occurs.)
Memory Address
(offset only)
Memory Value
(word)
007E
001D
007C
0240 (FLAGS)
007A
1266 (CS)
0078
0045 (IP)
0076
002C (DI)
0074
0063 (AX)
0072
0000 (BX)
c) (10 Points) At address 1266:0045 is a JZ instruction. Given the current state of
the registers and the contents of memory, determine if this jump will be taken or
not. Give the address of the instruction that will execute after the JZ instruction.
___1266:0047____________
The instruction in question is a JZ. The jump will be taken if the zero flag is
set. To determine if the zero flag will be set or not, it is necessary to look at
the last instruction executed which sets the zero flag. This is the CMP
instruction immediately prior to the JZ instruction. This instruction:
cmp al,[di]
compares the contents of AL, which is 63h, with the contents of memory
location 125A:002C (the value in DS:DI). This memory location contains
4Dh, which is not equal to 63h. The zero flag will not be set, and the jump
will not be taken. The next instruction to be executed would be at 1266:0047
EE-314
Spring 2001
Problem 6 Data)
You are stepping through the execution of an 8088 assembly language program. The
following information shows the state of the machine. Shown are memory dumps, a
disassembled listing of the part of the program that is currently executing, and the current
contents of the CPU registers.
Dump of Interrupt Vector Table:
0000:0000
0000:0010
0000:0020
0000:0030
BB
65
00
6A
08
04
00
00
0B
70
00
DA
02
00
C9
05
65
D7
28
82
04
04
00
00
70
00
DA
DA
00-16
C0-85
05-3A
05-9A
10
98
00
00
DB
00
DA
DA
05
F0
05
05
65
53
52
65
04
FF
00
04
70
00
DA
70
00
F0
05
00
....e.p.....e.p.
e.p.........S...
....(...:...R...
j...........e.p.
20
31
31
23
74
00
34
31
6F
57
20
00
20
65
4D
18
45
6C
69
00
45
63
64
00
33
6F
74
00
..Welcome to EE3
14 Exam #1.Welco
me to EE314 Midt
erm Exam #1.....
00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00-00 00 00 00 00 00 1D 00
................
................
Dump of the Program’s Data Segment:
125A:0000
125A:0010
125A:0020
125A:0030
00
31
6D
65
00
34
65
72
57
20
20
6D
65
45
74
20
6C
78
6F
45
63
61
20
78
6F
6D
45
61
6D-65
20-23
45-33
6D-20
Dump of the Program’s Stack Segment:
125E:0060
125E:0070
Listing of the program code:
1266:0035
1266:0036
1266:0037
1266:003A
1266:003C
1266:003F
1266:0041
1266:0043
1266:0045
1266:0047
1266:0049
1266:004C
1266:004E
1266:0051
1266:0053
1266:0055
46
47
803C00
7505
803D00
7412
8A04
3A05
74EE
7305
B8FFFF
EB07
B80100
EB02
33C0
C3
INC
INC
CMP
JNZ
CMP
JZ
MOV
CMP
JZ
JNB
MOV
JMP
MOV
JMP
XOR
RET
SI
DI
BYTE PTR [SI],00
0041
BYTE PTR [DI],00
0053
AL,[SI]
AL,[DI]
0035
004E
AX,FFFF
0055
AX,0001
0055
AX,AX
Current Contents of the CPU Registers:
AX=0063
DS=125A
BX=0000
ES=125A
CX=0018
SS=125E
DX=0000
CS=1266
SP=007E
IP=0043
BP=0000 SI=0013
FLAGS=0240
NMI Interrupt Service Routine:
NMIISR:
PUSH
PUSH
PUSH
CALL
POP
POP
POP
IRET
DI
AX
BX
HANDLENMI
BX
AX
DI
;process the NMI
DI=002C