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我必須制定一個計算器程序。我寫了這個：裝配程序中幾乎沒有錯誤。對於業餘愛好者
push ds
push 0000H
data segment
imsg1 db 13,10,'enter 1st number:$'
imsg2 db 13,10,'enter 2nd number:$'
imsg3 db 13,10,'sum:$'
imsg4 db 13,10,'diff:$'
imsg5 db 13,10,'div:$'
imsg6 db 13,10,'product:$'
num dw ?
num2 dw ?
data ends
MOV ax,data
MOV ds,ax
mov ah,09h
mov dx,offset imsg1
int 21h
mov cx,0
mov bx,10
mov num,0 ; Using num instead of dx, as we usually do, because we have to store another number too.
r1:mov ah,01h ; For storing first number
int 21h
cmp al,13
je yy
sub al,48
mov cl,al
mov ax,num
mul bx
add ax,cx
mov num,ax
jmp r1
yy:
;to enter 2nd number
mov ah,09h
mov dx,offset imsg2
int 21h
mov cx,0
mov bx,10
mov num2,0
r2:mov ah,01h
int 21h
cmp al,13
je yy1
sub al,48
mov cl,al
mov ax,num2
mul bx
add ax,cx
mov num2,ax
jmp r2
yy1:
;TO PRINT PROMPT MSG FOR SUM
mov ah,09h
mov dx,offset imsg3
int 21h
mov ax,num ; Move the numbers to registers before doing any operation
mov bx,num2 ; Because we will also need the numbers for other operations
add ax,bx ; Adding and storing in ax
mov cx,0
mov bx,10
mov dx,0
r3: ; To print the sum
mov dx,0
div bx
add dx,48
push dx
inc cx
cmp ax,0
jg r3
mov ah,02h
print:
pop dx
int 21h
loop print
;TO PRINT SUBTRACTION PROMPT
mov ah,09h
mov dx,offset imsg4
int 21h
mov ax,num
mov bx,num2
sub ax,bx ; Subtracting and storing in ax
mov cx,0
mov bx,10
mov dx,0
r4: ; Printing the subtracted value
mov dx,0
div bx
add dx,48
push dx
inc cx
cmp ax,0
jg r4
mov ah,02h
print1:
pop dx
int 21h
loop print1
;TO PRINT DIVISION PROMPT
mov ah,09h
mov dx,offset imsg5
int 21h
mov dx,0
mov ax,num
mov bx,num2
div bx ; Quotient stored in AX
mov cx,0
mov bx,10
mov dx,0
r5: ; Printing the Quotient
mov dx,0
div bx
add dx,48
push dx
inc cx
cmp ax,0
jg r5
mov ah,02h
print2:
pop dx
int 21h
loop print2
;TO PRINT MULTIPLICATION PROMPT
mov ah,09h
mov dx,offset imsg6
int 21h
mov dx,0
mov ax,num
mov bx,num2
mul bx ; Solution in AX
mov cx,0
mov bx,10
mov dx,0
r6: ; Printing the solution
mov dx,0
div bx
add dx,48
push dx
inc cx
cmp ax,0
jg r6
mov ah,02h
print3:
pop dx
int 21h
loop print3
ret
main endp
code ends
end main
而且我有錯誤：
line 5 (data segment) parser: instruction expected
line 18(mov dx,offset imsg1): comma or end of line expected
line 39(mov dx,offset imsg2): comma or end of line expected
line 62(mov dx,offset imsg3): comma or end of line expected
line 87(mov dx,offset imsg4): comma or end of line expected
line 112(mov dx,offset imsg5): comma or end of line expected
line 138(mov dx,offset imsg6): comma or end of line expected
line 162(main endp): error:parser:instruction expected
line 163(main ends): error:parser:instruction expected
line 164(end main): error:parser:instruction expected
我試圖解決這些問題，因爲沒有成功的一週。對不起，我已經把整個代碼放在這裏，但它不是那麼長的片段，所以也許有人可以幫助我.. 謝謝！
什麼彙編？看起來像Masm/Tasm代碼，錯誤消息看起來像他們可能來自Nasm ... –
只是一個給出未指定語法的猜測，但你可能需要冒號後imsg1：等等，讓他們被識別爲標籤？ –
我想在NASM中編譯它。添加冒號沒有奏效。我已經通過輸入segment .data解決了第一個錯誤（數據末尾也有錯誤，但我已經刪除了這一行） –