I need to create a program in Assembly that can count characters in one line and save the length of the shortest into a variable. But I don't really understand what to do and can't find help anywhere.
I have this, but it’s far from what I want it to do.
cpu 8086
segment code
start mov bx,data
mov ds,bx
mov bx,stack
mov ss,bx
mov sp,dno
radek mov ah,0x0a
mov dx,nacteno
int 21h
mov al,[nacteno+1]
cmp al,0
jne size
jz konec
size mov bl,[rslt34]
cmp al,bl
jle radek
mov [rslt34],al
jmp radek
konec hlt
segment data
nacteno db 255, ?
resb 255
rslt34 db 0
segment stack
resb 16
dno: db ?
that can count characters in one line
Because you're using the DOS.BufferedInput function 0Ah there's not much counting needed since you get it for free in the second byte of the structure. So be it.
Surprisingly perhaps, your attempt is rather good. But because of the way you initialize rslt34 (rslt34 db 0), the cmp al,bl jle radek will not close in on the shortest string but on the longest string.
You should setup rslt34 to 255 and use next code:
cmp al, 0 ; Exit on empty string
je konec
size:
cmp al, [rslt34]
jnb radek ; Ignore if not below what we already have
mov [rslt34], al ; Update with a length that is smaller than before
jmp radek
konec:
The length byte that DOS gives you must be treated as an unsigned number. Therefore don't use signed conditional jumps like jle and friends. Use the unsigned conditional jumps like jb (JumpIfBelow), jnb (JumpIfNotBelow), and others.
segment stack resb 16
Realistically, the BIOS/DOS stack should be at least 256 bytes, preferably 512 bytes.
konec hlt
The usual way to exit from a DOS program uses next instructions:
mov ax, 4C00h ; DOS.TerminateProgram
int 21h