EMU8086 将 32 位数字除以 16 位数字给出意外的 0 余数 [英] EMU8086 dividing 32 bit number by a 16 bit number gives unexpected 0 remainder

问题描述

我试图使用 emu8086 工具将(无符号)8A32F4D5 除以 C9A5.我预计商为 AF73H，余数为 94B6H.编写以下代码后，我得到了正确的商数，但余数变为 0000h.我错过了什么吗?

.MODEL 小.堆栈 100 小时.数据.代码主程序;初始化DS移动轴,@DATAMOV DS,AX;在此处输入您的代码MOV DX, 8A32HMOV AX, 0F4D5HMOV BX, 0C9A5HDIV BX;退出到DOSMOV AX,4C00HINT 21H主终端结束主要

EMU8086 中的输出:

解决方案

这看起来像是 EMU8086 中的一个错误.这个无符号除法没有被零除，也没有溢出(

如果使用 IDIV 指令对除法进行签名，它将产生除以零异常因为除法溢出.

I was trying to divide (Unsigned) 8A32F4D5 by C9A5 using emu8086 tool. I expected the quotient to be AF73H and the remainder be 94B6H. After writing the following code, I was getting correct quotient but the remainder became 0000h. Am I missing something?

.MODEL SMALL
.STACK 100H
.DATA 
.CODE 

MAIN PROC 
; initialize DS
MOV AX,@DATA 
MOV DS,AX 
; enter your code here
MOV DX, 8A32H
MOV AX, 0F4D5H 
MOV BX, 0C9A5H

DIV BX

;exit to DOS 
               
MOV AX,4C00H
INT 21H 

MAIN ENDP
    END MAIN 

The output in EMU8086:

解决方案

This looks like a bug in EMU8086. There is no division by zero nor is there an overflow with this unsigned division (DIV). You are correct that 0x8A32F4D5 divided by 0xC9A5 has a remainder of 0x94B6. To verify this I ran this code with Turbo Debugger in DOSBOX and got the expected results:

Had this been signed division using the IDIV instruction it would produce a division by zero exception because of division overflow.

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