1. High-level language structures
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2. Branching structures IF-THEN
IF condition is true
THEN
execute true-branch statements
End_if
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3. Branching structures IF-THEN
condition
false true
True branch
statements
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4. example Replace the number in AX by its absolute value.
Pseudo code algorithm is
IF AX <0
Then
replace AX by –AX
END_IF
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5. example Code: CMP AX,0 ;AX<0? JNL END_IF ;no, exit , JGE
NEG AX ;yes, change sign
END_IF:
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6. IF_THEN_ELSE IF condition is true THEN execute true-branch statements
execute FALSE-branch statements
End_IF
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7. IF_THEN_ELSE FALSE TRUE CONDITION FALSE BRANCH TRUE BRANCH STATEMENTS
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8. EXAMPLE
Suppose AL and BL contains extended ASCII characters. Display the one that comes first in the character sequence.
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9. example Pseudo code algorithm is IF AL < BL THEN
display the char in AL
ELSE
display the char in BL
END_IF
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10. EXAMPLE Code: MOV AH,2 CMP AL,BL ELSE_: MOV DL,BL JA ELSE_ DISPLAY:
MOV DL,AL
JMP DISPLAY
ELSE_: MOV DL,BL
DISPLAY:
INT 21H
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11. We use the unsigned jump because we are comparing extended characters.
JMP (unconditional jump) is needed to skip the false branch.
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12. CASE
A case is a multiway branch structure that tests a register, variable, or expression for particular values or range of values. The general form is as follows:
case expression
values_1: statements_1
values_2: statements_2
…………..
values_n: statements_n
END_CASE
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13. CASE EXPRESSION VALUE_1 VALUE_N VALUE_2 ……. STATEMENTS_1 STATEMENTS_2
STATEMENTS_N
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14. EXAMPLE
IF AX contains a negative number, put -1 in BX; if AX contains 0, put 0 in BX; if AX contains a positive number, put 1 in BX.
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15. example Pseudo code algorithm is Case AX < 0:put -1 in BX
END_CASE
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16. Example Code: CMP AX,0 ZERO: JL NEGATIVE MOV BX,0 JE ZERO JMP END_CASE
JG POSITIVE
NEGATIVE:
MOV BX,-1
JMP END_CASE
ZERO:
MOV BX,0
JMP END_CASE
POSITIVE:
MOV BX,1
END_CASE
NOTE: only one CMP is needed, because jump instructions do not affect the flags.
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17. example
If AL contains 1 or 3, display “o”; if AL contains 2 or 4, display “e”.
Pseudo code algorithm is
Case AL
1, 3: display “o”
2, 4: display “e”
End_case
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18. example Code: CMP AL,1 JE ODD ODD: CMP AL,3 MOV DL, ’o’ JMP DIS_
JE EVEN
CMP AL,4
JE EVEN
JMP END_
ODD:
MOV DL, ’o’
JMP DIS_
EVEN:
MOV DL, ’e’
DIS_:
MOV AH,2
INT 21H
END_:
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19. Branches with compound conditions
Sometimes the branching condition in an IF or CASE takes the form
Condition_1 and condition_2
Condition_1 or condition_2
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20. AND conditions
An AND condition is true if and only if condition_1 and condition_2 are both true.
Likewise, if either condition is false, then the whole thing is false.
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21. Example Read a character, and if it’s an uppercase letter, display it.
Pseudo code algorithm is
read a character (into AL)
If (‘A’ <= character) and (character <=‘Z’)
Then
Display character
End_if
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22. example Code: MOV ah,1 INT 21h MOV DL,AL CMP al,’A’ MOV AH,2 JL end_if
CMP al,’Z’
JG end_if
MOV DL,AL
MOV AH,2
INT 21H
End_if:
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23. OR conditions
condition_1 OR condition_2 is true if at least one of the conditions is true.
if both conditions are false, then the whole thing is false.
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24. example
Read a character. If it’s “y” or “Y”, display it; otherwise, terminate the program.
Pseudo code algorithm is
Read a character (into AL)
if (character = “y” or character= “Y”)
Then
Display character
Else
Terminate the program
End_if
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25. example THEN: MOV AH,1 MOV AH,2 INT 21h MOV DL,AL CMP AL, ’y’ INT 21H
JMP END_IF
ELSE_:
MOV AH,4CH
END_IF:
MOV AH,1
INT 21h
CMP AL, ’y’
JE THEN
CMP AL,’Y’
JMP ELSE_
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26. Looping structures
A loop is a sequence of instructions that are repeated.
For loop:
this is a loop structure in which the loop statements are repeated a known number of times (a count-controlled loop).
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27. For loop_count times do statements end_for
Initialize count
statements
For loop_count times do statements end_for
Count = count - 1
Count=0
False
True
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28. LOOP instruction
The LOOP instruction can be used to implement a FOR loop.
General form:
LOOP des_label
The counter for the loop is the register CX.
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29. LOOP instruction CX is initialized to loop_count.
Execution of the LOOP instruction causes CX to be decremented automatically.
If CX is not 0, control transfers to des_label.
If CX =0, the next instruction after LOOP is done.
Des_label must precede the LOOP instruction by no more than 126 bytes.
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30. LOOP instruction So, a FOR loop can be implemented as follows.
; initialize CX to loop_count
Des_label:
; body of the loop
loop des_label
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31. example Write a count-controlled loop to display a row of 80 stars.
Pseudo code algorithm is
For 80 times do
display ‘*’
end_for
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32. example MOV CX,80 MOV AH,2 MOV DL,’*’ TOP: INT 21H LOOP TOP 23/04/2019
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33. LOOP instruction A loop instruction is executed at least once.
If CX = 0 when the loop is entered, the loop instruction causes CX to be decremented to FFFFH.
So, the loop is executed more times.
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34. LOOP instruction Solution:
use the instruction JCXZ (jump if CX is zero) before the loop.
JCXZ des_label
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35. LOOP instruction A loop implemented as follows is bypassed if CX is 0.
JCXZ skip
top:
;body of the loop
loop top
Skip:
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36. While Loop This loop depends on a condition. Pseudo code
while condition do
statements
end_while
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37. While Loop
condition
True
statements
False
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38. example
Write some code to count the number of characters in an input line.
Pseudo code
initialize count to 0.
read a character
while character <> carriage_return do
count=count + 1
end_while
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39. example MOV DX,0 ; or bx, cx, ax?? MOV AH,1 INT 21H WHILE_: CMP AL,0DH
JE END_
INC DX
JMP WHILE_
END_ :
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40. Repeat loop Another conditional loop is the repeat loop. Pseudo code
statements
until condition
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41. Repeat loop
statements
condition
False
True
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42. example Write some code to read characters until a blank is read.
Pseudo code
Repeat
read a character
until character is a blank
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43. example MOV AH,1 REPEAT: INT 21H CMP AL,’ ’ JNE REPEAT 23/04/2019
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44. WHILE versus REPEAT
In a while loop, the loop can be bypassed if the terminating condition is initially false.
The statements in a repeat must be done at least once.
Code for a repeat loop is likely to be shorter because there is only a conditional jump at the end, but a while loop has two jumps: a conditional jump at the top and a jmp at the bottom.
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45. Programming with high-level structures
Top-down program design:
The original problem is solved by solving a series of sub problems, each of which is easier to solve than the original problem.
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46. example Prompt the user to enter a line of text.
On the next line, display the capital letter entered that comes first alphabetically and the one that comes last. If no capital letters are entered, display “No capital letter”.
sample execution:
type a line of text:
THE QUICK BROWN FOX JUMPED.
first capital = B last capital = X
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47. Top down program design
First refinement:
Display the opening message.
Read and process a line of text.
Display the results.
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48. example TITLE PGM6_2 : FIRST LAST CAPITALS .MODEL SMALL .STACK 100H
.DATA
PROMPT DB 'Type a line of text ',0dH,0AH,'$'
NOCAP_MSG DB 0DH,0AH,' No Capitals $'
CAP_MSG DB 0DH,0AH,'FIRST CAPITAL = '
FIRST DB '['
DB ' LAST CAPITAL = '
LAST DB $'
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49. Step 1 .CODE MAIN PROC ; initialize DS MOV AX,@DATA MOV DS, AX
; first step display opening message
MOV AH,9
LEA DX,PROMPT
INT 21H
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50. Step 2: read and process a line of text
Read a character.
While character is not a carriage return do
If character is a capital letter
Then
If character precedes first capital
First capital = character
End_if
If character follows last capital
Last capital = character
Read a character
End_while
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51. ;read and process a line of text MOV AH,1 INT 21H WHILE_:
; while character is not carriage ;return do
CMP AL,0dH
JE END_WHILE
; if character is a capital letter
CMP AL,'A'
JNGE END_IF
CMP AL,'Z'
JNLE END_IF
; then
; if character precedes FIRST ;capital
CMP AL,FIRST
JNL CHECK_LAST
; then FIRST CAPITAL = ;character
MOV FIRST,AL
; end_if
CHECK_LAST :
; if character follows last capital
CMP AL,LAST
JNG END_IF
; then LAST CAPITAL = character
MOV LAST,AL
;end_if
END_IF:
; read a character
INT 21H
JMP WHILE_
END_WHILE :
A=65, Z=90,
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52. Step 3: display the results
if no capitals were typed
then
display “no capitals”
Else
display first capital and last capital
End_if
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53. ; if no capital were typed CMP FIRST, '[' JNE CAPS ; then
; display results
MOV AH,9
; if no capital were typed
CMP FIRST, '['
JNE CAPS
; then
LEA DX,NOCAP_MSG
JMP DISPLAY
CAPS :
LEA DX,CAP_MSG
DISPLAY :
INT 21H
;end_if
; DOS exit
MOV AH,4CH
MAIN ENDP
END MAIN
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