emu8086 - Why does the following 8086 assembly code can only display numbers upto 2559? -
what trying ?
i want 16bit number user , print on console.
what problem ?
well code works fine til number entered less 2600 movement enter 2600 displays "40" , 2601 "41" , on. shouldn't display numbers upto 65535 ? because storing value in bx register 16bit , stores @ 65535. why 2559 ?
my code:
.model small .data msg db 10,13,'enter 16bit number: $' newline db 10,13,'$' .code main: mov ax, @data mov ds, ax mov cl, 10 mov bx, 0 mov ah, 09h lea dx, msg int 21h call get16bitnum mov ah, 09h lea dx, newline int 21h mov ax, '$' push ax mov ax, bx store: div cl cmp al, 0 mov bh, 0 mov bl, ah push bx je continue mov ah, 0 jmp store continue: pop ax cmp ax, '$' je ext mov bx, ax mov ah, 02h mov dx, bx add dx, 48 int 21h jmp continue ext: mov ah, 04ch int 21h proc get16bitnum aggregate: mov ah, 01h int 21h cmp al, 13 je return mov ah, 0 sub al, 48 mov dx, bx mov bx, ax mov ax, dx mul cl add bx,ax jmp aggregate return: ret endp end main
you don't retrieve 16-bit number!
you keep desired input in bx, , need multiply whole of bx 10. using word sized multiplication.
proc get16bitnum aggregate: mov ah, 01h int 21h cmp al, 13 je return mov ah, 0 sub al, 48 ;ax 0 9 xchg ax, bx ;new digit temporarily in bx mov cx, 10 mul cx ;product in dx:ax, won't use dx! add bx ,ax ;add new digit jmp aggregate return: ret you don't display 16-bit number
the procedure convert number text need use word sized division.
for excellent explanation on how see recent post displaying numbers dos. explains in great detail need know converting numbers. uses same technique of pushing value in stack (you used $ character this) know number ends.
ps. if find info in the linked post useful don't hesitate upvote it. (of course hope find answer useful too!)
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