Question

我需要帮助我的汇编程序中的第一个程序。我必须将用户输入的值从十进制转换为二进制。我不知道如何将值显示为小数，我接下来该怎么做。任何人都可以一步一步地指导下一步做什么。

    .model small
    .stack 100h`

    .data
            txt1 db "Enter binary value:" ,10,13, "$"
            txt2 db "BIN: " ,10,13, "$"


    .code

        main proc
        mov ax, @data
        mov ds, ax
        ;clear screen
        mov ah,0fh
        int 10h
        mov ah,0
        int 10h
        ;show first text
        mov ah, 9
        mov dx, offset txt1
        int 21h
        call Number


        main endp


        Number proc
        mov cx,5
        xor bx,bx

        read:
        mov ah,0
        int 16h
        cmp al,'0'
        jb read
        cmp al, '9'
        ja read
        mov ah,0eh
        int 10h
        loop read
        Number endp

        mov ax, 4c00h
        int 21h

        end main

Answer 1

我认为你会好的。

; Read an integer from the screen and display the int in binary format
; and continue until number is negative.
again:            ; For loop
    call read_int ; take the integer from screen
    cmp eax,0     ; look if number is not negative
        JL end:       ; if less than zero program ends.
    mov ecx,32    ; for loop we set ecx to 32 ; ATTENTION we not specified type. So compiler will get error.

    mov ebx,eax   ; we will lost our number in eax, so I take it to ebx
START:
    xor eax,eax   ; eax = 0
    SHL ebx,1     ; shift the top bit out of EBX into CF
    ADC eax,0     ; EAX  = EAX + CF + 0 ADD CARRY FLAG, so eax is zero we add zero. The new eax will exact value of Carry Flag which is out bit.
    call print_int ; Then we print the CF which we took the eax.
LOOP start:   ; Loop looks ecx if not 0 it goes start.  
call print_nl ; For next number we print a new line
JMP again:    ; For take new number

    END:      ; End of the program.

setc al也可以工作而不是adc eax,0，并且在某些CPU上效率更高。

Answer 2

不完全清楚你要做什么。我猜“十进制到二进制”，但提示“输入二进制值”。我认为这意味着一串“1”和“0”。我不会问他们的“小数”值 - 你会得到像“1.23”，你没有能力处理。只要问他们一个号码。也许“数字小于65536”，这可能是（？）你想要的。

警告！未经测试的代码！

    Number proc
    mov cx,5 ; loop counter?
    xor bx,bx ; "result so far"?


    read:
    mov ah,0
    int 16h

; wanna give 'em the option to enter
; less than the full five digits?
    cmp al, 13 ; carriage return
    jz finis

    cmp al,'0'
    jb read
    cmp al, '9'
    ja read
    mov ah,0eh
    int 10h
; Assuming al still holds your character...
    sub al, '0' ; convert character to number
    mov ah, 0 ; make sure upper byte is clear
    imul bx, bx, 10 ; multiply "result so far" by 10
    ; jc overflow ;  ignore for now
    add bx, ax ; add in the new digit
    ; jc overflow ; ignore for now

    loop read
finis:
; now our number is in bx
; it is conventional to return values in ax
    mov ax, bx

overflow: ; I'm just going to ignore it
    ; spank the user?
    ; go right to exit?
    ret ; maybe endp generates this. two shouldn't hurt
    Number endp

现在我想你要打印那个数字的二进制（“1”和“0”）表示......

    Printbin proc
; what does "proc" do? Do you know?
    mov bx, ax
    mov cx, 16 ; 16 bits to do, right?
    mov ah, 2
 top:
    mov dl, '0'
    shl bx, 1 ; shift leftmost bit to carry flag
    adc dl, 0 ; bump the "0" up to "1", if set
    int 21h
    loop top
    ret

    endp ; ?

自从我做了DOS之后已经有一段时间了，所以可能会有严重的错误，但它可能会给你一些想法。

在汇编程序中将十进制转换为二进制

2 个答案: