总装科 [英] Assembly Division

问题描述

在我的节目，一个十六进制数被除以10，其余的检查。

In my program, a hex number is divided by ten and the remainder is checked.

第一次分裂的表现不错;然而，第二次分裂后，程序出错。我是新来组装，而我找不到问题出在哪里？

First division is performed well; however, after the second division, the program goes wrong. I am new to assembly, and I couldn't find where the problem is...

下面是code段：

ORG 1000

    MOV AX, 0x04B4 (1204 decimal value )
    MOV BX, 0x000A ( 10 decimal value )
    MOV CX, 0x0000

    DIV BX ( After this part, AX is 120 decimal and DX 4 decimal )

    CMP DX, 0x0000
    JE eq1

    ADD CX, 0x0002
    JMP con1

    eq1:    ADD CX, 0x0001  

    con1:

    DIV BX ( But, after this division AX becomes 6677 ( 26231 decimal and DX remains 4 decimal )
    CMP DX, 0x0000

感谢您的帮助！

推荐答案

的 DIV BX 指令把在 32位在DX值：通过BX AX。既然你不进行初始化DX，被除数的高位字是什么垃圾留在从previous计算DX寄存器，所以你真的除以0x00040078 = 10。262314的结果是正确的：一个商数的26231与4的余数。

The DIV BX instruction divides the 32-bit value in DX:AX by BX. Since you're not initializing DX, the upper word of the dividend is whatever garbage was left in the DX register from the previous computation, so you're really dividing 0x00040078=262314 by 10. The result is correct: a quotient of 26231 with a remainder of 4.

在第一次分裂是一定是幸运，DX正好是0开始。

In the first division is must have been pure luck that DX happened to be 0 initially.

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