Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

26
Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

description

Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control. Chapter Outline Short, near and far address JMP Instruction The CMP Instruction Conditional Jump instruction The Loop instruction While Loop REPEAT Loop. Short,near,and far addresses. - PowerPoint PPT Presentation

Transcript of Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Page 1: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Lecture 5 and Lecture 6Chapter 7 –Program Logic and Control

Page 2: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Chapter Outline

Short, near and far address

JMP Instruction

The CMP Instruction

Conditional Jump instruction

The Loop instruction

While Loop

REPEAT Loop

Page 3: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Short,near,and far addresses

1- A short address, limited to a distance of -128 to 127 bytes

2- A near address, limited to a distance of -32,768 to 32,767 bytes

3- A far address, which may be within the same segment at a distance over 32K or in other segment

SHORT NEAR FAR

JMP YES YES YES

JXXX(conditional jump) YES YES NO

LOOP YES NO NO

CALL NlA YES YES

Page 4: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Unconditional Jumps - The JMP Instruction

• The JMP (jump) instruction causes an unconditional transfer of control (unconditional jump).

• Syntax: JMP destination

•Example

JMP L10……..

L10: INC CX

JMP SHORT/NEAR/FAR address

Page 5: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Unconditional Jumps - The JMP Instruction

•Backward and Forward jumps

Backward: L10: ……. JMP L10

Forward: JMP L10

……. L10:

Page 6: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

The CMP Instruction

• The jump condition is often provided by the CMP (compare) instruction

• Syntax: CMP destination, source

• Compares by computing destination contents minus source contents.

• The result is not stored, but the flags are affected.

• Destination may not be a constant.

• CMP is just like SUB, except that destination is not changed.

Page 7: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Conditional Jumps

• Syntax Jxxx destination_label

• Example JNZ PRINT_LOOP

• If the condition for the jump is true, the next instruction to be executed is the one at destinaltion_label (PRINT_LOOP), which may precede or follow the jump instruction itself.

• If the condition is false, the instruction immediately following the jump is done next.

• For JNZ, the cindition is that the result of the previous operation is not zero.

Page 8: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Conditional Jumps

• Signed Jumps: used for signed interpretations.

Symbol Description Condition for Jumps JG/JNLE jump if grater than ZF = 0 & SF = OF

jump if not less than or equal JGE/JNL jump if grater than or equal SF = OF

jump if not less than JL/JNGE jump if less than SF <> OF

jump if not greater than or equal JLE/JNG jump if less than or equal ZF = 1 or SF <> OF

jump if not grater than

Page 9: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Conditional Jumps

• Unsigned Jumps: used for unsigned interpretations.

Symbol Description Condition for Jumps JA/JNBE jump if above CF = 0 & ZF = 0

jump if not below or equal JAE/JNB jump if above or equal CF = 0

jump if not below JB/JNAE jump if below CF = 1

jump if not above or equal JBE/JNA jump if below or equal CF = 1 or ZF = 1

jump if not above

Page 10: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Conditional Jumps

• Single Flag Jumps: operates on settings of individual flags.

Symbol Description Condition for Jumps JE/JZ jump if equal/ jump if equal to 0 ZF = 1 JNE/JNZ jump if not equal/ jump if not 0 ZF = 0 JC jump if carry CF = 1 JNC jump if no carry CF = 0 JO jump if overflow OF = 1 JNO jump if no overflow OF = 0 JS jump if sign negative SF = 1 JNS jump if nonnegative sign SF = 0 JP/JPE jump if parity even PF = 1 JNP/JPO jump if parity odd PF = 0

Page 11: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

IF-THEN-ELSE

• Example: Suppose AL and BL contain extended ASCII characters. Display the one that comes first in the character sequence.

• Solution: Pseudocode:

IF AL <= BL THEN

display the character in AL ELSE

display the character in BLEND_IF

continue

Page 12: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

IF-THEN-ELSE

CMP AL, BL ; AL <= BL?JNBE ELSE_ ; no, display char in BL

; AL <= BLMOV DL, AL ; move char to be displayedJMP DISPLAY ; go to display

ELSE_: ; BL < ALMOV DL, BL

DISPLAY:MOV AH, 2 ; prepare to displayINT 21h ; display it

It can be coded as follows: ; if AL <= BL

; then

Page 13: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Branches with compound Conditions

• Sometimes the branching condition in an IF or CASE takes the form:

condition_1 AND condition_2 or condition_1 OR condition_2

where condition_1 and condition_2 are either true or false.

AND condition

OR condition

Page 14: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

AND Condition

• An AND condition is true if and only if all conditions are true.

• Example: Read a character, and if it’s an uppercase letter, display it.• Solution: Pseudocode:

Read a character (into AL)IF ('A' <= character) and (character <= 'Z') THEN

display characterEND_IF

continue

Page 15: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

AND Condition

It can be coded as follows: ; read a character

; if ('A' <= char) and (char <='Z')

; then display char

MOV AH,1 ; prepare to readINT 21h ; char in AL

CMP AL, 'A' ; char >= 'A'?JNGE END_IF ; no, exitCMP AL, 'Z' ; char <= 'Z'?JNLE END_IF ; no, exit

MOV DL, AL ; get charMOV AH, 2 ; prepare to displayINT 21h ; display char

END_IF:

Page 16: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

OR Condition

• An OR condition is true if at least one of the conditions is true.

• Example: Read a character. If it’s 'y' or 'Y', display it; otherwise, terminate the program.• Solution: Pseudocode:

Read a character (into AL)IF (character = 'y') or (character = 'Y') THEN

display character ELSE

terminate the programEND_IF continue

Page 17: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

OR Condition

It can be coded as follows: ; read a character

; if (char = 'y') or (char = 'Y')MOV AH,1 ; prepare to readINT 21h ; char in AL

CMP AL, 'y' ; char = 'y'?JE THEN ; yes, go to display itCMP AL, 'Y' ; char = 'Y'?JE THEN ; yes, go to display itJMP ELSE_ ; no, terminate

THEN:MOV DL, AL ; get charMOV AH, 2 ; prepare to displayINT 21h ; display charJMP END_IF ; and exit

ELSE_:MOV AH, 4ChINT 21h ; DOS exit

END_IF:

Page 18: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Loop Instruction

• The LOOP instruction can be used to implement a for loop.

• Syntax:

• The counter for the loop is the register CX, which is initialized to loop_count.

• Execution of the LOOP instruction causes CX to be decremented automatically.

• If (CX < > 0) control transfers to destination_label else the next instruction after LOOP is done.

LOOP SHORT address

Page 19: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

Loop Instruction

• Using the instruction LOOP, a FOR loop can be implemented as follows:

; initialize CX to loop_count TOP:

; body of the loopLOOP TOP

Page 20: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

FOR Loop

• Example: Write some code to display a row of 80 stars.

• Solution: Pseudocode:

FOR 80 times DO display '*'END_IF

It can be coded as follows:MOV CX, 80MOV AH, 2MOV DL, '*'

TOP:INT 21hLOOP TOP

; what if CX =0?

MOV CX, 80MOV AH, 2MOV DL, '*'JCXZ SKIP ;jump if CX=0

TOP:INT 21hLOOP TOP

SKIP:

Page 21: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

WHILE Loop

• This loop depends on a condition.

• Pseudocode: WHILE condition DO

statements END_WHILE

Page 22: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

WHILE Loop

• Example: Write some code to count the number of characters in an input line.

• Solution: Pseudocode:

initialize count to 0read a characterWHILE character <> carriage_return DO count = count + 1

read characterEND_WHILE

continue

Page 23: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

WHILE Loop

It can be coded as follows:

WHILE_:

END_WHILE:

MOV DX, 0 ; DX counts charactersMOV AH, 1 ; prepare to readINT 21h ; character in AL

CMP AL, 0Dh ; CR?JE END_WHILE ; yes, exitINC DX ; not CR, increment countINT 21h ; read a characterJMP WHILE_ ; loop back

Page 24: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

REPEAT Loop

• This loop depends on a condition.

• Pseudocode: REPEATStatementsUNTIL conditions

Page 25: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

REPEAT Loop

• Example: write code to read characters until a blank is read

• Pseudocode: REPEATRead characterUNTIL character is blank The code is:

MOV AH,1REAPEAT: INT 21H CMP AL,’ ‘ JNE REAPEAT

Page 26: Lecture 5 and Lecture 6 Chapter 7 –Program Logic and Control

WHILE Versus REPEAT

• Use of a WHILE loop or a REPEAT loop is a matter of personal preference.

• A WHILE loop can be bypasses if the terminating condition is initially false. (a REPEAT loop must be done at least once)

• The code for a REPEAT loop is likely to be a little shorter because there is only one jump. (WHILE loops has two jumps)