3) Write a 8086 assembly program that:
- Load AL by an arbitrary number,
- Check whether it is an odd/even number,
- If it is an odd number, it will be converted to its next larger even number and will be shown in port 0 in the trainer kit.
- If it is an even number, it will be converted to its previous smaller odd number and will be shown in port 0 in the trainer kit.
Assuming that the input number is provided in the AX register, the following program can perform the given task:
; Initialize data segment
MOV AX, @data
MOV DS, AX
; Load AL by input number
MOV AL, AH
; Check odd/even
AND AL, 00000001B ; check the least significant bit
JZ is_even ; if it is zero, the number is even
; Convert odd to even
INC AL ; add 1 to make it even
JMP show_result
is_even:
; Convert even to odd
DEC AL ; subtract 1 to make it odd
show_result:
; Output the result to port 0
MOV DX, 0
MOV AH, 02h ; select output mode for DX
OUT DX, AL ; send the result to port 0
; Terminate program
MOV AH, 4Ch
INT 21h
The program starts by initializing the data segment to the location where the code and data segments are stored in memory. Then, it loads the input number from the AX register into the AL register.
Next, the program checks whether the number is odd or even by ANDing the least significant bit (LSB) of AL with a bit mask that has all bits set to 0 except the LSB. If the result of the AND operation is zero, the number is even and the program branches to the is_even label. Otherwise, it proceeds to the next instruction, which converts the odd number to the next larger even number by incrementing it.
If the number is already even, the program jumps to the show_result label directly after the LSB check and subtracts 1 from AL to make it the previous smaller odd number.
In either case, the program proceeds to the show_result label, where it sets the DX register to 0 to select port 0 and uses the OUT instruction to send the result in AL to that port.
Finally, the program terminates by setting the AH register to 4Ch (the exit function code) and calling interrupt 21h to exit to the operating system.
Sure! Here's a step-by-step breakdown of the 8086 assembly program:
Step 1: Define data segment and stack segment
```assembly
.data ; Data segment
input db 0 ; Define input variable to store the arbitrary number
.code ; Code segment
.startup ; Entry point of the program
mov ax, @data ; Initialize data segment register
mov ds, ax ; Point DS to data segment
mov ax, 0 ; Initialize AX register with 0
mov bx, 0 ; Initialize BX register with 0
push bx ; Push BX onto the stack
push ax ; Push AX onto the stack
jmp start ; Jump to start label
```
Step 2: Read input and check for even/odd
```assembly
start:
lea dx, input ; Load address of input variable into DX
mov ah, 01 ; Set AH to 01 to read a character
int 21h ; Interrupt 21h to read a character from console
sub al, 30h ; Convert ASCII character to decimal
xor ah, ah ; Clear AH register
test al, 01h ; Test the least significant bit to check for even/odd
jz even ; Jump to the even label if number is even
inc al ; If odd, increment the number to obtain next even number
jmp output ; Jump to output label to show the result on port 0
even:
dec al ; If even, decrement the number to obtain previous odd number
output:
mov dx, 0 ; Initialize DX register with 0
mov al, 01h ; Set AL to 01h to indicate output to port 0
out dx, al ; Output the result to port 0
pop ax ; Pop AX from the stack
pop bx ; Pop BX from the stack
mov ah, 4ch ; Set AH to 4Ch to terminate the program
int 21h ; Interrupt 21h to terminate the program
```
Step 3: Terminate the program
```assembly
.exit ; End of program
end ; End of program
```
Please note that this program assumes the usage of an 8086 Trainer Kit with a display port at port 0 for showing the result. You may need to adapt it to your specific hardware setup.