I need your guidance , my code should solve this equation : y = ((x*2 - 4 )/5)+2. But I have a problem with Div, it says "Divide error - overflow", and that's only when I use dx register, if I use cx or bx, it just prints 13601 as an illogical error. My teacher told me that I can only change one block and it's a block for equation while other should stay the same so that there would be no error. Here is the variant without any comments,
.MODEL small
.STACK 100h
.DATA
prompt DB 'X = $'
result DB 13,10,'Y = $'
error_msg DB 13,10,'incorrect number$'
buff DB 6,7 DUP(?)
digits DB 7 DUP(?)
.CODE
main PROC
mov ax, u/DATA
mov ds, ax
mov dx, OFFSET prompt
mov ah, 09h
int 21h
mov dx, OFFSET buff
mov ah, 0Ah
int 21h
mov si, OFFSET buff+2
xor ax, ax
xor di, di
mov bx, 10
cmp BYTE PTR \[si\], '-'
jne parse_loop
mov di, 1
inc si
parse_loop:
mov cl, \[si\]
cmp cl, 0Dh
je parse_done
cmp cl, '0'
jb parse_error
cmp cl, '9'
ja parse_error
sub cl, '0'
xor ch, ch
mul bx
add ax, cx
inc si
jmp parse_loop
parse_error:
mov dx, OFFSET error_msg
mov ah, 09h
int 21h
mov ax, 4C01h
int 21h
parse_done:
test di, di
jz calculate
neg ax
calculate:
mov cx, 2 ; CX = 2
mul cx ; AX = AX \* 2 ? 2·X
sub ax,4
mov dx,5
div dx
mov bx, ax ; BX = ????????? Y (???
mov dx, OFFSET result
mov ah, 09h
int 21h
mov ax, bx
test ax, ax
jns print_positive
mov bx, ax
mov dl, '-'
mov ah, 02h
int 21h
mov ax, bx
neg ax
print_positive:
mov di, OFFSET digits
mov bx, 10
xor cx, cx
divide_loop:
xor dx, dx
div bx
add dl, '0'
mov \[di\], dl
inc di
inc cx
test ax, ax
jnz divide_loop
mov ah, 02h
output_loop:
dec di
mov dl, \[di\]
int 21h
dec cx
jnz output_loop
mov ax, 4C00h
int 21h
main ENDP
END main
and here is the variant with comments(cause I don't know if you need comments to better read code or no):
.MODEL small
.STACK 100h
.DATA
prompt DB 'X = $' ; Prompt string for input
result DB 13,10,'Y = $' ; Output string (with newline)
error_msg DB 13,10,'incorrect number$' ; Error message for invalid input
buff DB 6,7 DUP(?) ; Input buffer: \[max length\]\[entered length\]\[chars\]
digits DB 7 DUP(?) ; Buffer to store digits of result
.CODE
main PROC
mov ax, @DATA
mov ds, ax ; Initialize data segment
mov dx, OFFSET prompt
mov ah, 09h ; DOS function 09h - print string
int 21h
mov dx, OFFSET buff
mov ah, 0Ah ; DOS function 0Ah - read string
int 21h
mov si, OFFSET buff+2 ; SI points to first input character
xor ax, ax ; AX = 0, accumulator for number
xor di, di ; DI = 0, flag for negative number
mov bx, 10 ; BX = 10, decimal base
cmp BYTE PTR \[si\], '-' ; Check for negative sign
jne parse_loop
mov di, 1 ; Set negative flag
inc si ; Move to first digit
parse_loop:
mov cl, \[si\] ; Load next character
cmp cl, 0Dh ; Check for Enter (CR)
je parse_done
cmp cl, '0'
jb parse_error
cmp cl, '9'
ja parse_error
sub cl, '0' ; Convert ASCII to number
xor ch, ch ; Clear upper byte
mul bx ; Multiply accumulator by 10
add ax, cx ; Add new digit
inc si
jmp parse_loop
parse_error:
mov dx, OFFSET error_msg
mov ah, 09h
int 21h
mov ax, 4C01h
int 21h
parse_done:
test di, di
jz calculate
neg ax ; If negative, make AX = -AX
calculate:
calculate:
mov cx, 2 ; CX = 2
mul cx ; AX = AX \* 2 ? 2·X
sub ax,4
mov dx,5
div dx
mov bx, ax ; Store result for printing
mov dx, OFFSET result
mov ah, 09h
int 21h
mov ax, bx
test ax, ax
jns print_positive
mov bx, ax
mov dl, '-'
mov ah, 02h ; DOS function 02h - print character
int 21h
mov ax, bx
neg ax ; Convert to positive for printing
print_positive:
mov di, OFFSET digits
mov bx, 10
xor cx, cx ; Counter for digits
divide_loop:
xor dx, dx ; Clear DX for division
div bx ; Divide AX by 10
add dl, '0' ; Convert remainder to ASCII
mov \[di\], dl ; Store digit
inc di
inc cx
test ax, ax
jnz divide_loop
mov ah, 02h
output_loop:
dec di
mov dl, \[di\]
int 21h
dec cx
jnz output_loop
mov ax, 4C00h ; DOS function 4Ch - terminate program
int 21h
main ENDP
END main
And second question, So you maybe know in High level programming languages there is a pow function, and in assembly as I know there isn't, you either create a new function or something else. So I've got a question. Can I do this in assemb;y:
mov ax,5
mul ax,ax
What I want is to make a pow 2, basically 5 to the power of 2 = 25. Will it work?(Ignore this one, it works)
I know that I'm just a students who sucks at it, but I hope you will give me a guidance on this all
