如何添加/ sub / div / mul超过4个值？ 汇编语言

Question

very newbie question, how can I express arithmetic equations using more than 4 values?

(14×3) + 16/4-3

 ORG 0 MOV AL, E MOV BL, 3 MUL AL, BL ; MOV CL, 10 MOV DL, 4 DIV CL, DL ; ADD AL, CL MOV ??, 03 <--- what to put, DL is the last register SUB AL, ?? <--- what to do END 

Answer 1

First, MUL and DIV only take 1 argument. Search for 'intel mul' and 'intel div' to see the instruction details:

For 8 bits:

Using an 8 bit register r8 as argument (where r8 is one of the 16 8 bit registers),

• MUL r8 wil multiply r8 with al and store the result in ax . This is because, for instance, multiplying 127 with 127 is bigger than 8 bits (but never more than 16).
• Similarly, div r8 will divide ax by r8 , put the result in al , and the remainder in ah .

For a 16 bit argument:

• MUL r16 will multiply r16 (a 16 bit register) with ax , and store the result in dx:ax , ie, the high word in dx , and the low word in ax .

• Again similarly, DIV r16 will divide dx:ax by r16 , put the result in ax and the remainder in dx .

Your calculation

Calculating 14×3 + 16/4 - 3 is done like this:

    ; First term: 14x3
    mov al, 14
    mov bl, 3
    mul bl        ; ax = 42 (or, al=42 and ah=0)


    ; next we're going to DIV, which uses `AX`, so we better copy our result out of ax:
    mov cx, ax    ; or: mov cl, al


    ; Second term, 16/4 (16/3 would be more interesting!)
    mov ax, 16    ; we could mov al, 16 since ah is already 0
    mov dl, 4
    div dl        ; now al=4 and ah=0  (With 16/3, al=5 and ah=1!)


    ; Add to our result:
    add cl, al
;   adc ch, 0     ; take care of overflow if we want a 16 bit result


    ; calculate the 3rd term    
    mov al, 3     ; that was easy :-)

    ; and add the 3rd term to our result:
    sub cl, al    ; we could have done sub cl, 3

I hope you get the idea!