我怎样才能从10个最简单的方法中赚取10000,1000,100 [英] How can i make 10000,1000,100 from 10 the easiest way

问题描述

我正在寻找上述问题的解决方案. 我需要生成100,1000和10000(十进制数). 因为整个练习都是为了计算:

i am looking for a solution for the question abowe. I need to generate 100,1000 and 10000(decimal numbers). Because the whole exercise is to caluclate:

10000 * X + 1000 * Y + 100 * Y + 10 * V + 1 * C

10000*X+1000*Y+100*Y+10*V+1*C

我知道我可以通过mul命令来完成，但是在那种情况下 我必须在堆栈上做很多工作.

I know i can do it by the mul command,but in that case i have to work a lot with stack.

我当时在想某种转变，但不知道该怎么做 带有十进制数字.

I was thinking some kind of Shifting, but don't know how to do it with decimal numbers.

带有二进制数

mov al,2h;mov al,10b;
shl al,1

Preferebla是一种恶劣的环境.感谢您的帮助

Preferebla is masm enviroment.Thank you for your help

Update1: 我不想使用mul或imul(少于32位数字) 我有一个数字，例如string(db)可以说24575，我有伪代码，但是还没写完

Update1: I dont want to use mul nor imul(less then 32 bit number) I have a number as string(db) lets say 24575,i have hier the pseudo code,not finished

    mov cx,5
    Calc:
mov di,offset first
add di,2;so we are on the adress of "2"


;ASCI number is 2 and i want to get 20000
    mov al,10d

    ;than ASCI number is 4 and i want to get 4000

    ;than ASCI number is 5 and i want to get 500

    ;than ASCI number is 7 and i want to get 70

    ;than ASCI number is 5 and i want to get 5


inc di
    loop Calc

first:db"A$"
number :db"24575$"

推荐答案

要计算10000*X+1000*Y+100*Z+10*V+1*C，请使用转换10*(10*(10*(10*X+Y)+Z)+V)+C.这样，您每次迭代只需乘以10.正如我在评论中所写，可以通过使用10=8+2来避免使用mul.因此，您的代码可能看起来像这样:

To calculate 10000*X+1000*Y+100*Z+10*V+1*C use the transformation 10*(10*(10*(10*X+Y)+Z)+V)+C. This way you only need to multiply by 10 in every iteration. As I have written in my comment, you can avoid a mul by using 10=8+2. As such your code may look something like this:

    mov cx, 5               ; number of digits
    mov di, offset number   ; pointer to digits
    xor ax, ax              ; zero result
next:
    mov dx, ax              ; save a copy of current result
    shl ax, 3               ; multiply by 8
    add ax, dx              ; 9
    add ax, dx              ; 10
    movzx dx, byte ptr [di] ; fetch digit
    sub dx, '0'             ; convert from ascii
    add ax, dx              ; add to sum
    inc di                  ; next digit
    loop next

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