solution manual to ASSEMBLY LANGUAGE PROGRAMMING AND ORGANIZATION OF THE IBM PC

1. A solution manual to
Assembly language
Programming
and the organization of
the IBM PC
Chapter 6
Flow control instructions
BY:- Ytha Yu & Charles Marut
prepared by :
Warda Aziz
Wardaaziz555@gmail.com

2. Question 1:
Write assembly code for each of the following decision structures.
A) IF AX<0
THEN
PUT -1 IN BX
END_IF
SOLUTION:
;this is just an instruction, editing is required while implementing
CMP AX, 0 ;condition if(ax<0)
JL loop_ ;goto loop
JMP end_if
Loop_: ;loop:
MOV BX,-1 ;loop body
end_if: ;else {}
;DOS EXIT ;return 0
B) IF AL<0
THEN
Put FFh in AH
ELSE
Put 0 in AH
END_IF
SOLUTION:
;this is just an instruction, editing is required while implementing
CMP AL,0
JL loop_
JMP loop1
Loop_:
MOV AH, 0FFh
JMP end_if
Loop1:
MOV AH, 0
JMP end_if
end_if:
;DOS EXIT
C) Suppose DL contains the ASCII code of a character.
(IF DL>= “A”) AND (DL<= “Z”)
THEN
Display DL
END_IF
Solution:
No editing required, just add the format of assembly
MOV AH,1
INT 21h
MOV dl, al

3. CMP dl,"A"
JGE jump
JMP exit
jump:
CMP DL, 'Z'
JLE jumpe
JMP exit
jumpe:
MOV AH,2
INT 21H
JMP exit
exit:
;DOS exit
D) if AX<BX
THEN
IF BX<CX
THEN
PUT 0 IN AX
ELSE
Put 0 in BX
End_ifEnd_if
Solution:
MOV AX,5
MOV BX,6
MOV CX,7
CMP AX,BX ;if ax is less then bx , then check another condition
JB label
JMP end_if
MOV AX, 0
Label:
CMP BX, CX ;if BX<CX satisfies then mov 0 in ax and terminate the code
JB label2
MOV BX,0
JMP end_if
Label2:
MOV AX,0
JMP end_if
end_if:
MOV ah,4ch
INT 21h
E) if(AX<BX) OR (BX<CX)
Then
Put 0 in DX
Else
Put 1in DX
End_if

4. Solution:
No editing required
MOV AX,7
MOV BX,6
MOV CX,5
CMP AX,BX ;if ax is less then bx , mov 0 in dx
JB label
CMP BX, CX ; OR if BX<CX satisfies then mov 0 in dx
JB label
MOV DX,1 ;else mov 1 in dx
JMP end_if
Label:
MOV DX,0
JMP end_if
end_if:
MOV AH,4ch
INT 21h
F) if AX<BX
THEN
Put 0 in AX
Else
If(BX<CX)
Then
Put 0 in BX
Else
Put 0 in cx
end_if
End_if
Solution:
MOV AX,7
MOV BX,6
MOV CX,5
CMP AX,BX ;if ax is less then bx , mov 0 in ax
JB label
CMP BX, CX ;else if ax is greater then bx then perform this section
JB label2 ;if bx<cx , mov 0 in bx
MOV CX,0 ;else mov 0 in cx
JMP end_if
label: ;label
MOV AX , 0

5. JMP end_if
Label2: ;label 2
MOV BX,0
JMP end_if
end_if:
MOV AH,4Ch
INT 21h
Question 2:
Use a CASE structure to code the following:
 Read a chracter.
 If it’s ‘A’, then execute carriage return
 If it’s ‘B’, then execute line feed
 If it’s any other chracter , then return to DOS
ANS:
MOV AH, 1
INT 21h
MOV BL,AL ;reading a character and storing in another register
CMP Bl, 'A' ;IF character = A
JE carriage ;jump carriage
CMP Bl,'B' ;else if character=B
JE line ;jump to line
JMP exit ;else exit
carriage:
MOV AH,2
MOV DL, 0Dh
INT 21h
MOV DL, BL
INT 21H
JMP exit
line:
MOV AH,2
MOV DL, 0Ah
INT 21H
MOV DL, BL
INT 21H
exit:
MOV AH,4ch
INT 21h
Question 3:
Write a sequence of instruction to code each of the following
A) Put the sum 1+4+7+…+148 in AX
B) Put the sum 100+95+90+…+5 in AX
Solution:
A)

6. ;Put the sum 1+4+7+...+148 in AX main endp
MOV CX, 49;counter for FOR LOOP
MOV AX,1 ;adds the sum of series and store the result
MOV BX,1 ;used to move the series 3 steps forward forward
Label:
ADD AX,BX
ADD BX,3
LOOP Label
;DOS exit
INT 21H
MOV AH,4CH
INT 21H
B) Put the sum 100+95+90+…+5 in AX
MOV CX, 19;counter for FOR LOOP
MOV AX,100 ;adds the sum of series and store the result
MOV BX,100 ;used to move the series 3 steps forward forward
Label:
ADD AX,BX
SUB BX,3
LOOP Label
;DOS exit
INT 21H
MOV AH,4CH
INT 21H
Question 4:
Employee LOOP to do the following:
 Put the sum of the first 50 terms of the arithmetic sequence
1,5,9,13,… in DX
 Read a character and display it 80 times on the next line
 Read a five character password and overprint it by executing
a carriage return and display five X’s. you need not store the
input characters anywhere.
Put the sum of the first 50 terms of the arithmetic sequence 1,5,9,13,… in DX
MOV CX,50
MOV BX,1
MOV DX,1
label:
ADD DX,BX
ADD BX, 4
LOOP label
Read a character and display it 80 times on the next line
MOV CX, 80
MOV AH,1

7. INT 21h
MOV AH,2
MOV DL,AL
label:
INT 21H
LOOP label
Read a five character password and overprint it by executing a carriage return
and display five X’s. you need not store the input characters anywhere.
MOV AH,7
INT 21H
INT 21H
INT 21H
INT 21H
INT 21H
MOV AH,2
MOV DL,0DH
INT 21H
MOV CX,5
LABEL:
MOV DL,'X'
INT 21H
LOOP LABEL
Question 5:
The following algorithm may be used to carry out division of two
non-negative numbers by repeated subtraction
;let AX contains the quotient
;BX contains dividend
;DX contains divisor then
MOV AX,0
LABEL:
CMP BX,DX
JGE label2
JMP end
LABEL2:
INC AX
SUB DX,BX
JMP label
EXIT:
;DOS exit
Question 6:
The following algorithm may be used to carry out multiplication
of two positive numbers M and N by repeated addition:
Initialize product to 0
Repeat
Add m to product
Decrement N
Untill n=0

8. Write a sequence of instructions to multiplay AX by BX and put
the product in CX. You may ignore the possibility of overflow
Let ax contains the product
MOV CX,0
LABEL:
ADD CX,AX
DEC BX
CMP BX,0
JNE label
Question 7:
It is possible to setup a count controlled loop that will continue to
execute as long as some condition is satisfied. Using LOOPE, LOOPZ,
LOOPNE, LOOPNZ do the following :
a) Write instructions to read characters until either a non blank
character is typed , or 80 characters have been typed. Use
LOOPE
b) Write instructions to read characters until either a carriage
return is typed , or 80 characters have been typed. Use LOOPNE
Write instructions to read characters until either a non blank character is typed ,
or 80 characters have been typed. Use LOOPE
MOV CX,80
MOV AH,1
Label:
INT 21h
CMP AL, 20h
LOOPE EXIT
LOOP LABEL
EXIT:
MOV AH,4CH
INT 21H
Write instructions to read characters until either a carriage return is typed , or
80 characters have been typed. Use LOOPNE
MOV CX,80
MOV AH,1
Label:
INT 21h
CMP AL, 0Dh
LOOPNE LABEL
EXIT:
MOV AH,4CH
INT 21H
Question 8:
Write a program to display a “?”, read two capital letters, and
display them on the next line in alphabetical order.

9. CODE:
.MODEL SMALL
.STACK 100h
.DATA
ms db 'Enter two letters $'
ch1 DB ?
ch2 DB ?
.CODE
MAIN PROC
MOV AX,@data
MOV DS, AX
;prompt the user
MOV AH, 9
LEA DX, ms
INT 21h
; read two characters
MOV AH, 1
INT 21h
MOV ch1,AL
INT 21h
MOV ch2, AL
;comparing the values
CMP ch1, AL
JB ascending
JA descending
JMP exit
ascending:
MOV AH,2
MOV DL, ch1
INT 21h
MOV DL, ch2
INT 21H
JMP exit
descending:
MOV AH,2
MOV DL, ch2
INT 21h
MOV DL, ch1
INT 21H
JMP exit
exit:
MOV ah,4ch
INT 21h
MAIN ENDP
END MAIN

10. Output:
Question 9:
Write a program to display the extended ASCII characters
(ASCII codes 80h to FFh). display 10 characters per line, separated
by blanks. Stop after the extended characters have been displayed
once.
Solution:
.MODEL SMALL
.STACK 100h
.DATA
.CODE
MAIN PROC
;prompt the user
MOV AH, 2
MOV CL, 0 ;for space
MOV DL, 80H ;initializing dl and bl to character at address 80h
MOV BL,80h
WHILE_:
MOV DL,BL ;restoring in dl for displaying
INT 21H
CMP DL, 0FFh
JE exit
INC Dl
MOV BL,DL ;temporarily storing dl
MOV DL,20h ;for space
INT 21h
INC CL ;incrementing in char for space
CMP cl, 10
JE line
MOV DL,BL
JMP WHILE_
line:
MOV dl,0ah
INT 21h
MOV dl,0dh
INT 21h
MOV cl,0
JMP while_
exit:
MOV ah,4ch

11. INT 21h
MAIN ENDP
END MAIN
Output
Question 10:
Write a program that will prompt the user to enter a hex digit
character (“0”, … , “9” or “A”, … , “”F”), display it on the next line
in decimal and ask the user if he want to do it again. If the user types
“y” or “Y”, the program repeats; if the user types anything else, the
program terminates, if the user enters an illegal character, prompt
the user to try again.
Solution:
CODE:
.MODEL SMALL
.STACK 100h
.DATA
msg DB 0AH,0DH,'Enter a hex character from ‘0’,…, ‘9’ or ‘A’,…, ‘F’$'
ch1 DB ?
ch2 DB ?
msg1 DB 0AH,0DH,"in decimal it is "
ch3 DB ?, "$"
msg2 DB 0AH,0DH,"In decimal it is: 1"
C2 DB ?,'$ '
msg3 DB 0AH,0DH,"do you want to check more results?",0ah,0dh,"type y for
yes , n for no "
c3 DB ?, "$"
msg4 DB 0AH,0DH,"invlaid input try again $"
.CODE
MAIN PROC
MOV AX,@DATA
MOV DS, AX
again:
;prompt the user

12. MOV AH, 9
LEA dx, msg
INT 21h
; read a character
MOV AH, 1
INT 21h
MOV ch1,AL
;CHECKING the hex character
CMP ch1, 39h ;39h=9
JBE dig
CMP ch1, 46h ;46H=F
JBE character
MOV AH,9
LEA DX, msg4
INT 21h
JMP again
input:
;if user want to take next input
MOV AH,9
LEA DX,msg3
INT 21h
MOV AH,1
INT 21h
MOV c3, AL
CMP c3, 'y'
JE again
CMP c3, 'Y'
JE again
JMP exit
character:
;CALCULATION for hex character
SUB AL, 11h
MOV C2,AL
MOV AH,9
LEA DX, msg2
INT 21H
JMP input
dig:
MOV AH,9
LEA DX, msg1
INT 21h

13. MOV AH,2
MOV DL, ch1
INT 21H
JMP input
exit:
MOV ah,4ch
INT 21h
MAIN ENDP
END MAIN
OUTPUT:
Question 11:
Do programming exercise 10, except that if the user fails to enter
a hex-digit character in three tries, display a message and terminates
the program.
CODE:
.MODEL SMALL
.STACK 100h
.DATA
msg DB 0AH,0DH,'Enter a hex character $'
ch1 DB ?
ch2 DB ?
msg1 DB 0AH,0DH,"in decimal it is "
ch3 DB ?, "$"
msg2 DB 0AH,0DH,"In decimal it is: 1"
C2 DB ?,'$ '
msg3 DB 0AH,0DH,"do you want to check more results?",0ah,0dh,"type y
for yes , n for no "
c3 DB ?, "$"

14. msg4 DB 0AH,0DH,"invalid input! $"
msg5 DB 0AH,0DH ,"Entered invalid input for three times! $"
.CODE
MAIN PROC
MOV AX,@DATA
MOV DS, AX
MOV BL,0
again:
;prompt the user
MOV AH, 9
LEA DX, msg
INT 21h
; read a character
MOV AH, 1
INT 21h
MOV ch1,AL
;CHECKING the hex character
CMP ch1, 39h ;39h=9
JBE dig
CMP ch1, 46h ;46H=F
JBE character
MOV AH,9
LEA DX, msg4 ;invalid input try again
INT 21h
INC bl
CMP BL,3
JE DISP
JMP again
DISP:
MOV AH,9
LEA DX, msg5
INT 21h
JMP exit
input:
;if user want to take next input
MOV AH,9
LEA DX,msg3
INT 21h
MOV AH,1
INT 21h
MOV c3, AL
CMP c3, 'y'
JE again
CMP c3, 'Y'
JE again

15. JMP exit
character:
;CALCULATION for hex character
SUB AL, 11h
MOV C2,AL
MOV AH,9
LEA DX, msg2
INT 21H
JMP input
dig:
MOV AH,9
LEA DX, msg1
INT 21h
MOV AH,2
MOV DL, ch1
INT 21H
JMP input
exit:
MOV ah,4ch
INT 21h
MAIN ENDP
END MAIN
OUTPUT:
Question 12:
Write a program that reads a string of capital letters , ending
with carriage return and display the longest sequence of consecutive
alphabetically increasing capital letters read.
CODE:
#copied from internet
.STACK 100h
.MODEL small
.DATA
inputmsg DB 'Enter a string of capital etters: $'
outputmsg DB 0dh, 0Ah, 'The longest consecutive increasing string is: $ '
.CODE

16. MAIN PROC
MOV AX, @DATA
MOV DS,AX
MOV CH,0 ;ch
MOV AH,9 ;dislaying input message
LEA DX,inputmsg
INT 21H
INPUT_1:
MOV AH,1
INT 21H
CMP AL, 0DH
JE end_
MOV CL,1
MOV BL,AL
MOV DH, AL
INPUT_2:
INT 21H
CMP AL,0dh
JE END_
INC bl
CMP BL,AL
JNE init
INC cl
JMP input_2
INIT:
CMP CH, CL
JL update
MOV CL,1
MOV BL,AL
MOV DH,AL
JMP input_2
update:
MOV ch,cl
MOV bh,dh
MOV cl,1
MOV bl,al
MOV dh,al
MOV input_2
end_:
CMP CH,CL
JL Reupdate
JMP end_2
reupdate:
MOV CH,CL
MOV BH,DH
JMP end_2
end_2:
MOV ah,9

17. Lea dx, outputmsg
INT 21h
MOV ah,2
MOV dl, bh
output:
CMP ch, 0
JE finish
DEC ch
INT 21h
INC dl
JMP output
finish:
MOV ah,4ch
INT 21h
main endp
end main
Output: