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你好,我有一个问题,我需要从用户那里得到输入,输入是一个数字和一个又一个数字,这个数字可以是一个双字

[英]hello , i got a question that i need to get input from user, the input is a number and digit after digit ,the number can be a doubleWord

I am making a program in which I want to take up to a 10 digit number我正在制作一个程序,我想在其中使用最多 10 位数字
(4,294,967,296) digit by digit from user and store all of the digits and make (4,294,967,296) 逐位来自用户并存储所有数字并制作
one number of them in EAX. EAX 中的其中一个。
for example: input = 1 2 3 4, EAX = 1234例如:输入 = 1 2 3 4,EAX = 1234
i tried with 2 digits for the start and i got some problems and i dont know how我试着用 2 位数字开始,但我遇到了一些问题,我不知道怎么做
to continue from here.从这里继续。 I would really appreciate your help, thanks in advance!!非常感谢您的帮助,在此先感谢!!

.model small
.STACK 100h
.data  
num dd ?
ten DB 10

.code
.386
   start:
   MOV AX, @DATA
   MOV DS, AX
   MOV AH,1
   INT 21H
   SUB AL,30H
   MOV BH,AL
   MOV AH,1
   INT 21H
   SUB AL,30H
   MOV CH,AL
   
   MOV AL,BL
   MUL ten 
   ADD AL,CH 
   push eax
   call printNum
   MOV AX, 4c00h
   INT 21h
   END start
MOV AL,BL MUL ten ADD AL,CH

The above would produce the value of your 2-digit number if you would have used the correct register.如果您使用了正确的寄存器,以上将产生您的 2 位数字的值。 You have stored the first digit in BH , but you're using BL here!您已将第一个数字存储在BH中,但您在这里使用的是BL

 .386 push eax

You are in emu8086.你在emu8086。 Forget about using 32-bit registers.忘记使用 32 位寄存器。 If you want to work with 32-bit numbers, you'll have to store them in a couple of 16-bit registers.如果要使用 32 位数字,则必须将它们存储在几个 16 位寄存器中。 eg DX:AX where DX holds the Most Significand Word and AX holds the Least Significand Word.例如DX:AX ,其中DX保存最高有效位字, AX保存最低有效位字。


To solve your task of building a 10 digit number up to 4GB-1, next codez will be useful:为了解决您构建高达 4GB-1 的 10 位数字的任务,下一个 codez 将很有用:

Multiplying the dword in DI:SI by 10DI:SI中的 dword 乘以 10

mov  ax, 10
mul  si         ; LSW * 10 -> DX:AX
mov  si, ax
xchg di, dx
mov  ax, 10
mul  dx         ; MSW * 10 -> DX:AX
jc   Overflow
add  di, ax
jc   Overflow

Adding the new digit AL=[0,9] to the dword in DI:SI将新数字AL=[0,9]添加到DI:SI中的 dword

mov  ah, ah
add  si, ax
adc  di, 0
jc   Overflow

Looping until the user presses Enter循环直到用户按下Enter

Input:
  mov  ah, 01h    ; DOS.GetKeyWithEcho
  int  21h        ; -> AL
  cmp  al, 13     ; Is it <ENTER> ?
  je   Done
  sub  al, '0'    ; From character to digit
  cmp  al, 9
  ja   Invalid

  ...

  jmp  Input
Done:

It's important that you decide how you will handle an invalid user input and what you will do for an input that yields a number bigger than 4GB-1.决定如何处理无效的用户输入以及如何处理产生大于 4GB-1 的数字的输入非常重要。 Many programs have been written that overlook this and then at some point mysteriously fail...许多程序都忽略了这一点,然后在某些时候神秘地失败了......

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