x86 Assembly:除法浮点异常除以11 [英] x86 Assembly: Division Floating Point Exception dividing by 11

问题描述

我正在尝试将859091除以11以获得商和余数，但是我在网上遇到了浮点异常:

I'm trying to divide 859091 by 11 to obtain the quotient and the remainder, but I'm getting Floating Point Exception on line:

div bx

这是我的 SASM 代码:

%include "io.inc"
section .data
  dividend dd 859091
  divisor  dw 11

section .text
global CMAIN
CMAIN:
  push ebp
  mov ebp, esp

  xor eax, eax
  xor ebx, ebx
  xor edx, edx

  mov ax, word [dividend]
  mov dx, word [dividend + 2]
  mov bx, word [divisor]    
  test bx, bx
  jz exit

  div bx

exit:
  leave
  ret

推荐答案

由于商不适合16位整数，因此您将获得除法溢出.

You're getting divide overflow because the quotient doesn't fit within a 16 bit integer.

您可以将红利分成上半部分和下半部分，以产生最多32位的商和16位的余数. dx = 0000 : ax = upper_dividend / divisor的余数成为第二除法的第二个红利的上半部分，因此第二个除法计算dx = remainder : ax = lower_dividend / divisor，这两个均不能溢出，因为余数严格小于除数.此过程可以扩展为更长的股息和商，每个单词和商数一个步骤，而每个除法步骤的其余部分将成为下一步部分股息的上半部分.

You can split up the dividend into upper and lower halves to produce up to a 32 bit quotient and 16 bit remainder. The remainder of dx = 0000 : ax = upper_dividend / divisor becomes the upper half of 2nd dividend for the 2nd division, so the 2nd division calculates dx = remainder : ax = lower_dividend / divisor, neither of which can't overflow because the remainder is strictly less than the divisor. This process can be extended for longer dividends and quotients, one step per word of dividend and quotient, with the remainder of each divide step becoming the upper half of the partial dividend for the next step.

使用MASM语法的示例:

Example using MASM syntax:

dvnd    dd 859091
dvsr    dw 11
;       ...
;       bx:ax will end up = quotient of dvnd/dvsr, dx = remainder
        mov     di,dvsr
        xor     dx,dx
        mov     ax,word ptr dvnd+2      ;ax = upr dvnd
        div     di                      ;ax = upr quot, dx = rmdr
        mov     bx,ax                   ;bx = upr quot
        mov     ax,word ptr dvnd        ;ax = lwr dvnd
        div     di                      ;ax = lwr quot, dx = rmdr

四字词示例:

dvnd    dq 0123456789abcdefh
dvsr    dw 012h
quot    dq 0
rmdr    dw 0
;       ...
        mov     di,dvsr
        xor     dx,dx                   ;dx = 1st upr half dvnd = 0

        mov     ax,word ptr dvnd+6      ;ax = 1st lwr half dvnd
        div     di                      ;ax = 1st quot, dx = rmdr = 2nd upr half dvnd
        mov     word ptr quot+6,ax

        mov     ax,word ptr dvnd+4      ;ax = 2nd lwr half dvnd
        div     di                      ;ax = 2nd quot, dx = rmdr = 3rd upr half dvnd
        mov     word ptr quot+4,ax

        mov     ax,word ptr dvnd+2      ;ax = 3rd lwr half dvnd
        div     di                      ;ax = 3rd quot, dx = rmdr = 4th upr half dvnd
        mov     word ptr quot+2,ax

        mov     ax,word ptr dvnd        ;ax = 4th lwr half dvnd
        div     di                      ;ax = 4th quot, dx = rmdr
        mov     word ptr quot,ax

        mov     rmdr,dx

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