没有div的ASM 8086分区 [英] ASM 8086 division without div

问题描述

我需要在asm 8086中编写类似b = a/6的程序，但没有DIV指令.我知道如何使用SAR，但是只有2,4,8,16 ...

I need to write in asm 8086 a program like b=a/6 but without the DIV instruction. I know how to do it with SAR but only 2,4,8,16...

mov ax,a
sar ax,1 ;//div a by 2
mov b,ax

我的问题是如何将其除以6?

my question is how can I do it to div by 6?

推荐答案

给出另一个答案的方法很简单强制循环，对于a的较大值可能需要一段时间.此版本使用较大的块(类似于长除法问题)，专门编码以将有符号数除以6:

The approach given an another answer is simple brute force loop, and can take a while for large values of a. This is a version that uses larger chunks (working it like a long division problem) specifically coded to divide a signed number by 6:

; signed divide by 6
    mov ax,a
    mov cx,1000h  ; initial count of how many divisors into ax to check for
    mov bx,6000h  ; value of "divisor * cx"
    xor dx,dx     ; result
top:
    cmp ax,bx
    jl skip
    ; we can fit "cx" copies of the divisor into ax, so tally them
    add dx,cx
    sub ax,bx
    ; optionally can have a "jz done" here to break out of the loop
skip:
    shr bx,1
    shr cx,1
    jnz top

    ; copy result into ax
    mov ax,dx

如果需要除6以外的值，则需要调整初始cx和bx值. cx是设置位14的除数的2的幂次幂(因为位15是符号位；对于无符号除法，您希望设置位15). bx是2的幂.如果a的初始值有限制，则可以调整cx和bx的初始值，但必须小心，因为如果您输入错误，将会得到错误的答案.使它们过小.

If you need to divide something other than 6, the initial cx and bx values need to be adjusted. cx is the power-of-two multiple of the divisor that leaves bit 14 set (since bit 15 is the sign bit; for an unsigned divide you'd want to have bit 15 set instead). bx is that power of 2. If there are limits on the initial value for a you can adjust the initial cx and bx values, but have to be careful because you'll get an incorrect answer if you make them too small.

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