Answer :
We need to update the value of the DL register in the program to display only the English alphabet without the additional characters. By changing the initial value to 41h (the ASCII value for 'A') and adjusting the comparison condition in the loop, we can display the desired output.
The modified Assembly program would look as follows:
.data
; no variables to declare
.code
; the code for the program
main proc
mov dl, 41h ; sets DL to 'A'
LP1:
mov ah, 2 ; display character
mov dl, al ; move value of AL to DL (ASCII code)
int 21h ; library call
inc dl ; increment DL (move to next character)
cmp dl, 5Bh ; compare DL with 5Bh ('[')
jns LP1 ; jump if not signed (DL is less than 5Bh)
mov ax, 4c00h ; ends the program
int 21h ; library call
main endp
end main
In this modified program, we initialize the DL register with the value 41h, which corresponds to the ASCII code of 'A'. Then, we enter a loop labeled LP1, where we display the character represented by the current value of DL using the mov ah, 2 instruction followed by int 21h (an interrupt call). After displaying the character, we increment the value of DL by one using the inc dl instruction.
To ensure that only the English alphabet is displayed, we modify the comparison condition in the loop. The new condition cmp dl, 5Bh compares DL with the ASCII code of '[' (5Bh). If DL exceeds this value, the jump instruction jns LP1 jumps back to the LP1 label, continuing the loop. This effectively skips the unwanted characters in the ASCII table.
Once the loop reaches the character represented by 'z', DL will exceed the ASCII code of '[', and the program will proceed to the termination part. The mov ax, 4c00h instruction terminates the program, and int 21h is used to make the corresponding interrupt call.
By modifying the program in this way, it will display only the English alphabet as requested.
Learn more about DL register here:
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