汇编语言中十进制到二进制的转换

在将十进制值转换为二进制时,我在分析这段代码时遇到了困难。谁能帮我解释一下下面的几行文字吗?顺便说一下,这是一个工作代码。我非常感谢你的帮助,这是为了学习。谢谢你!

calc :
    mul multiplier
    mov bl, byte ptr [si]
    mov bh, 00h
    add ax, bx
    inc si
    loop calc

    mov si, offset buf4 + 2
    mov bx, ax
    mov dx, 0000h
    mov ax, 8000h
convert :
    mov cx, 0000h
conv :
    cmp bx, ax
    jb cont3
    sub bx, ax
    inc cx
    jmp conv
cont3 :
    add cl, 30h
    mov byte ptr [si], cl
    inc si
    mov cx, 0002h
    div cx
    cmp ax, 0000h
    jnz convert

    mov byte ptr [si], '$'
    prnstr buf3
    prnstr buf4+2
stop :
    mov ax, 4c00h
    int 21h

代码的第一部分很容易理解(接受用户输入并检查其是否为有效的十进制数)。我已经添加了一些评论。。代码的前半部分如下所示:

prnstr macro msg     ;acts like a function to print string
    mov ah, 09h
    lea dx, msg
    int 21h
    endm

data segment
    buf1 db "Enter a decimal number : $"
    buf2 db 0ah, "Invalid Decimal Number...$"
    buf3 db 0ah, "Equivalent Binary number is : $"
    buf4 db 6
         db 0
         db 6 dup(0)
    multiplier db 0ah
data ends

code segment
    assume cs:code, ds:data
    start :
    mov ax,@data
    mov ds,ax
    mov es, ax   

    prnstr buf1         ;Display

    mov ah, 0ah         ;Get line function
    lea dx, buf4        ;adress of input buffer
    int 21h

    mov si, offset buf1 + 2 ;Load pointer to beginning of structure
    mov cl, byte ptr [si-1] ;determine end of loop?
    mov ch, 00h
    subtract :
    mov al, byte ptr [si]   ;Load 1 byte of buf1 to cl
    cmp al, 30h             ;check if not below 0
    jnb cont1

    jmp stop
 cont1 :
    cmp al, 3ah         ;check if not above 9
    jb cont2
    prnstr buf2         ;Display
    jmp stop
 cont2 :
    sub al, 30h             ;convert ASCII to Decimal number
    mov byte ptr [si], al   ;place bcd form back to pointed character
    prnstr buf2             ;Display
    inc si                  ;next character
    loop subtract           ;repeat until end of string
    mov si, offset buf1 + 2 ;Load pointer to beginning of structure
    mov cl, byte ptr [si-1] ;reset to determine end of loop using si
    mov ch, 00h
    mov ax, 0000h           ;reset ax to 0

    ;...CONTINUE TO calc.....