将负数与正数相除 [英] Divide a negative with a positive

问题描述

假设以下代码：

mov eax, 0
sub eax, 2

xor edx, edx
idiv base

其中 base 等于8（定义为 const int base = 8 ）。

where base equals to 8 (defined as const int base = 8).

因此，我希望在 eax 中得到0，在 edx 中得到2（或−2）。但是，划分的结果是一个很大的长期数字。很有可能是由 eax 注册表溢出引起的，当我从中减去2时。我认为 eax 被 idiv 操作解释为 unsigned int 。如何正确执行此除法？

As a result, I expect to get 0 in eax and 2 (or −2) in edx. However, the result of the division is a big long number. Most likely, it is caused by overflow in eax registry, when I subtract 2 from it. I think eax is interpreted by idiv operation as unsigned int. How do I perform this division properly?

我在x64系统上的Visual Studio 2013中编写代码。

I code in Visual Studio 2013 on x64 system.

推荐答案

通常的原因： edx 应该是64位股息的上半部分。因此，如果您有一个负数，则 edx 不能为零（至少必须设置符号位，否则它不是负数），并且您有32位符号数字，您必须将其符号扩展到 edx：eax 。

The usual reason: edx is supposed to be the "upper half" of a 64bit dividend. So if you have a negative number, edx can not be zero (at least the sign bit must be set, otherwise it's just not negative), and if you had a 32bit signed number, you have to sign-extend it into edx:eax.

cdq 指令确实如此，所以很容易：

Which is what the cdq instruction does, so it's easy:

mov eax, -2

cdq
idiv base

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