大会师的x86与DX：AX [英] Assembly x86 division with DX:AX

问题描述

我与MASM工作，我以前遇到过的情景，我不容易理解如何解决，例如：

I'm working with masm and I've encountered a scenario I do not readily understand how to solve, for example:

X = (A)/(C*D)

如果我多C *Ð第一，我的价值是DX：AX据我所知，我不能使用，作为除数。如果我做单独的司A / C和A / D，我冒丢失precision（从提醒等）的风险。什么是实现这个的最佳方式？

If I multiple C*D first, my value is DX:AX and to my knowledge, I cannot use that as a divisor. If I do division separately as A/C and A/D, I run the risk of losing precision (from the reminders, etc.). What's the best way to implement this?

推荐答案

当你正确地指出，你不能用一个32位的数字为16位除法除数，但由于我们只关心这样做整数除法，这不是一个问题。

As you correctly note, you can't use a 32-bit number as the divisor in a 16-bit division, but since we're only interested in doing integer division that's not a problem.

有两种情况考虑（无符号司）：

There are two cases to consider (for unsigned division):

• DX == 0 ： C *ð的结果符合16位，所以我们可以继续正常使用斧为16位除数。


• DX＆GT; （<！code> DX = 0 ）0 ： C *ð大于65335（大于 0xFFFF的）和 A ，这个数字将始终为0，其余的仅仅是 A 。



• DX == 0: The result of C*D fits in 16 bits so we can proceed as normal using ax as the 16-bit divisor.
• DX > 0 (DX != 0): C*D is greater than 65335 (0xFFFF) and the 16-bit unsigned division of A and that number will always be 0 and the remainder is simply A.

或者你可以做C和只是假设， C的结果* D 16位相符。 ：）

Or you could do as C and just assume that the result of C*D fits in 16 bit. :)

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