如何将字符串转换为Tasm中的数字？

Question

我做了一个程序，读取一个字符串并将其转换为数字。在十进制系统中输入的数字的字符串表示形式。转换的结果在寄存器ax中。 我使用tasm16。

我该怎么做，我可以转换一个字符串更大（超过65535）？

model small 
stack 100h 
printstring macro msg 
    mov ah,09h 
    mov dx,offset msg 
    int 21h 
endm 
data segment 
    cr equ 0dh 
    lf equ 0ah 
    errorInput db 'Wrong input',cr,lf,'$' 
    inputAmsg db 'Please input A',cr,lf,'$' 
    outstring db 'Output:',cr,lf,'$' 
    maxAbuf db 10 
     db ? 
     db 10 dup(0) 
     db '$' 
    result dw 0 

data ends 
code segment 
    assume cs:code, ds:data 
start: 
    mov ax,data 
    mov ds,ax 
;input A 
    printstring inputAmsg 
    mov ah,0ah 
    lea dx,maxAbuf 
    int 21h 
    printstring outstring 

    mov ah,09h 
    lea dx,maxAbuf+2 
    int 21h 
;in: 
;dx:address of string 
;out 
;dx:ax - result ;work only ax 
; 
    call Convert 
    jmp exit 
;in: 
;dx:address of string 
;out 
;dx:ax - result 
;9999989d = 989675h 
Convert proc 
     xor ax,ax 
     xor cx,cx 
     mov bx,10 ;base 
     mov si, offset dx 
     xor dx,dx 
getnewchar: mov cl,[si] 
     cmp cl,0dh ;if `$` exit 
     jz endproc 
     sub cl,'0' ;or 30h 
     mul bx 
     add ax,cx 
;I do not know what to write here, if there is an exit for the bit grid. 
     adc dx,0 
     inc si 
     jmp getnewchar 
endproc: 
    ret 
    endp 

exit: mov ax,4c00h 
    int 21h 
code ends 
end start 

Answer 1

你可以，例如，由10从16位到32位扩展您的乘法是这样的：

; file: mul32ten.asm 
; compile: nasm.exe mul32ten.asm -f bin -o mul32ten.com 

bits 16 
org 0x100 

    xor  dx, dx 
    xor  ax, ax 

    mov  si, MyStr 

NextDigit: 
    mov  bl, [si] 
    or  bl, bl 
    je  Done 
    call MulDxAxBy10 
    sub  bl, '0' 
    add  al, bl 
    adc  ah, 0 
    adc  dx, 0 
    inc  si 
    jmp  NextDigit 

Done: 
    ; dx:ax should be equal to 7FFFFFFF now 
    ret 

MulDxAxBy10: 
    push si 
    push di 
    shld dx, ax, 1 
    shl  ax, 1  ; dx:ax = n * 2 
    mov  si, ax 
    mov  di, dx  ; di:si = n * 2 
    shld dx, ax, 2 
    shl  ax, 2  ; dx:ax = n * 8 
    add  ax, si 
    adc  dx, di  ; dx:ax = n * 8 + n * 2 = n * (8 + 2) = n * 10 
    pop  di 
    pop  si 
    ret 

MyStr db "2147483647", 0