8086 microprocessor Programming (Part 7)

JUMP INSTRUCTIONS

OBJECTIVES

• To become more familiar with the 8086 compare instruction.
• To study the 8086 instructions for implementing conditional logic structures.

INTRODUCTION:

Here are no conditional logic structures in the 8086 assembly language. But we can implement almost any structure using a combination of statements. A logic structure is simply a group of conditional statements working together. The WHILE structure, for instance, has an implied IF statement at its beginning that evaluates a condition:

Structured Unstructured

DO WHILE; L1 IF (a>=b)THEN

<statement-1> jump to L2

<statement-2>   ENDIF

<statement-1>

ENDDO <statement-2>

:

jump to L1

L2: (continue here)

The angle brackets (<>) around the word statement identify a program statement. Two steps are involved in executing IF statement. First, an arithmetic or comparison instruction sets one or more flages based on its result. Second, a conditional jump instruction may cause the CPU to jump to a new address. The instructions used belong to one of two groups:

Group 1: Comparison and arithmetic instructions, in which the CPU sets individual flags according to the result.

Group 2: Conditional jump instructions, in which the CPU takes action based on the flags.

The instructions from the two groups work in tandem: An instruction from Group 1 is executed affecting the flags, then a conditional jump instruction from Group 2 executes, based on the value of one of the flags. In the following example the JZ instruction jumps to ‘next’ if AL = 0.

Cmp a1 , 0

Jz next

  next :               ;  label

Conditional Jump Instructions: A Conditional jump instruction, transfers control to a destination address when a flag condition is true. The syntax is:

J cond destination

Here cond refers to a flag condition, identifying the state of one or more flags. For example:

C Carry flag set

NC Carry flag not set (clear)

Z Zero flag set

NZ Zero flag not set (clear)

We have already seen that flags are set by arithmetic, comparison, and Boolean instructions. Each conditional jump instruction checks one or more flags returning a result of true or false. If the result is true the jump is taken; otherwise, the program does nothing and continues to the next instruction. Note that these instructions themselves do not affect any of the six status flags.

EXPERIMENT: Here are four parts of this experiment, each containing a relatively smaller program. It will be convenient for user to put the main body of the program in the program template defined earlier in experiment 4. This will help you save time writing four times the common MASM directives and initialization instructions.

PART-1: The following program is taken from chapter 7 of [1]. It converts each character type at the keyboard is uppercase. The program ends when ENTER is pressed. Write it in a text editor , assemble and link it and then run it from DOS prompt to verify.

Page, 132

Title  uppercase display program

dosseg

.model small

.stack 100h

carr_ret equ 0Dh ; ASCII code of carriage return code

main proc

label _ 1

mov ah. 08h ; input a character, no echo

int 21h

cmp al, carr_ret ; ENTER pressed?

Je label _ 3 ; yes: quit

cmp al, ‘a’ ; character < ‘a’?

Jb label _ 2 ; if below, display it

cmp al,   ‘z’ ; character > ‘z’ ?

ja label _ 2 ;  if above, display it

sub al , 20h ; no , subtract 20h from ASCII code

label _ 2

mov ah, 02h ; function display character

mov dl , al ; character is in Dl

int 21h ; call DOS interrupt

;  get another character

jmp label _ 1 ; unconditional jump

label _ 3

mov ; ax , 4c00h

int 21h

main endp

data ; variables here

end main

1. What happens if you enter a character other than from a-z?
2. If you want to make a program which converts an uppercase characters in to lower case, what changes will be required in the above program?
3. Which flags are tested the execution of ‘je’ , ‘jb’ , and ‘ja’ instructions. Use DEBUG to find the answer.

Part – II: The following program is adopted from chapter 7 of [1]. It finds the largest and smallest signed numbers in an array of integers. The numbers input don’t have any suffix (e,g, h, b, or d). This means that they will be treated as decimal numbers. These numbers will be converted by the MASM in to proper hex numbers. A negative number will be converted in to its 2’s complement from by the MASM. Run the program in DEBUG.

page, 132

title Largest and Smallest Signed Numbers

dosseg

.model small

.stack 100h

.code

num_digits equ 6

main proc

mov ax . @data ; initialize the DS register

mov ds , ax

mov di , offset  array

mov ax , [di]

mov largest . ax ; initialize largest

mov smallest . ax ; initialize smallest

mov cx , num digits-1

label_1

add di , 2h ; point to next number

mov ax , [di] ; get next number

cmp ax , smallest ; [DI] >= smallest?

jge label_2 ; yes : skip

mov smallest , ax ;no:  move  [DI] to smallest

label_2:

cmp ax , largest ; [DI]  <=  largest?

jle label_3 ; yes : skip

mov largest . ax ; no:  move [DI] to largest

label_3:

loop  label_1 ; repeat until  CX =0

mov ax , 4C00h

int 21h

main endp

 data ; variable here

array dw -1 , 2000, 32767, 500,0

largest dw ?

smallest dw ?

end main

1. Note down from DEBUG the hex equivalent of the above six numbers.
2. Modify the program, so that it finds the largest and smallest of the following array of numbers: 200, -300, -7699, -3289, -2278, -9910, -5436, 98, 10.
3. Why are the contents of CX registers initialized at num_digits – 1 and not num_digits?
4. What is the range of integers over which this program works ? What happens if the array has a number out of this range.

Part –III:

• write a program that sorts, in descending order, a given list of integers in the range of

-32, 768  to +32,767. First make a flow chart with the help of following algorithm:

Call the elements of the second number, A(0), A(1), A(2), …., A(n). Take the first number in array, which is A(0), and compare it with the second number, A(1),. If A(0) is greater than A(1), the two numbers are swapped otherwise they are left alone. Next A(0) is compared with A(2) and on the basis of the result of this comparison, they are either swapped or left alone. This sequence is repeated until A(0) has been compared with all numbers up through A(n). When this is complete, the smallest number will be in A(0) position. New A(1) must be compared with A(2) through A(n) in the same way . After this is done, the second smallest number will be in A(1) position. The same procedure is repeated for A(2) through A(n-1) to complete the soft.

REFERENCES:

1. Kip R. Irvine. Assembly Language for the IBM-PC Macmillan Publishing Company, 1990.
2. Avtar Singh and Walter A. Triebel, The 8086 and 80286 Microprocessors, Hardware, Software, and Interfacing. Prentice-Hall Inc, 1990.