It reads the integers just fine and finds the smallest one just fine as well. However, it gives wrong results for sums > 9.
I am trying to calculate the sum in BH as a decimal. However if I for example ask it to add 9 and 9 it stocks 12 in BH instead of 18.
data segment
msg db "chiffre: $"
petit db "plus petit chiffre: $"
somme db "somme: $"
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov es, ax
mov ah, 09h
mov dl, offset msg
int 21h
mov ah, 01h
int 21h
mov bl, al ; bl contains the smallest number
mov bh,al
sub bh,'0'
mov cx,3
etq:
mov ah, 02h
mov dl,0Dh
int 21h
mov ah, 02h
mov dl,0Ah
int 21h
mov ah, 09h
mov dl, offset msg
int 21h
mov ah, 01h
int 21h
cmp al, bl
jb min
jae reste ; jump to the end of the loop
min:
mov bl, al
reste:
sub al,'0'
add bh,al
loop etq
The debugger is showing the correct result but does so using hexadecimals.
Displaying numbers with DOS explains how you can convert your number into a string of decimal characters that you can then display on the screen, so you don't have to interpret the hexadecimal values found in the debug window.
mov dl, offset msg
This is an error that you should be aware of. The DOS.PrintString function 09h expects the address of the string in the word-sized DX register. You should not only load the low byte DL and trust the high byte DH to contain 0. So write mov dx, OFFSET msg.
cmp al, bl jb min jae reste ; jump to the end of the loop min: mov bl, al reste:
You can easily avoid having to jump over the assignment of the new minimum. If only you would select the opposite conditional jump, you can replace the pair jb min jae reste by the single instruction jnb reste.
I see that your program is repeating some instructions. I the below code snippet I have the loop do 4 iterations after a suitable initialization of BX before the loop begins:
mov bx, 000Ah ; BH is sum (0), BL is smallest number (10)
mov cx, 4
etiquette:
mov ah, 09h
mov dx, OFFSET msg
int 21h
mov ah, 01h
int 21h ; -> AL = ['0','9']
sub al, '0' ; From ['0','9'] to [0,9]
cmp al, bl ; (*)
jnb passer
mov bl, al ; Assign a new minimum
passer:
add bh, al ; Sum
mov ah, 02h
mov dl, 13
int 21h
mov dl, 10
int 21h
loop etiquette
(*) Because that smallest number variable BL was initialized at 10 which is greater than any (decimal) digit the user can input, the very first cmp al, bl will already set the below condition and therefore the very first new minimum will get assigned from the very first user input (just like it was in your program).
The biggest sum that your program can expect is 36 (9+9+9+9). As an alternative approach, and instead of using the methods described in the Q/A for which I provided a link above, you could use next code to display a small number in the range [0,99]:
mov al, bh ; Sum eg. 18
aam ; -> AH = AL / 10 AL = AL % 10 eg. AH = 1 AL = 8
add ax, '00' ; Convert into text eg. AH = '1' AL = '8'
mov cx, ax
mov ah, 02h
mov dl, ch ; Tens eg. '1'
int 21h
mov dl, cl ; Ones eg. '8'
int 21h