如何检查寄存器是否可以被 7 整除? [英] How can I check if a register is evenly divisible by 7?

问题描述

我想检查寄存器 ax 的值是否可以被 7 整除，余数 = 0.我该怎么做?

I want to check if the value of the register ax is divisible by 7, with remainder = 0. How can I do it?

推荐答案

您可以使用与此答案中所示相同的方法，当然有不同的常数.

You can use the same way as shown in this answer, with different constants of course.

7 mod 2¹⁶ 的模乘逆是 0x6db7，这(根据定义)意味着 x = n * 7 形式的数字将服从 x * 0x6db7 = n，其中 n 将小于或等于 0xffff/7.所以如果你从 x * 0x6db7 中得到更大的东西，你知道它不是 7 的倍数.此外，非 7 的倍数也不能映射到低结果，因为乘以一个奇数模 a二的幂是双射的.

The modular multiplicative inverse of 7 mod 2¹⁶ is 0x6db7, which (by definition) means that numbers of the form x = n * 7 will obey x * 0x6db7 = n, where n will be less than or equal to 0xffff / 7. So if you get something bigger out of x * 0x6db7, you know it's not a multiple of 7. Also, non-multiples of 7 cannot also map to low results, because multiplication by an odd number modulo a power of two is bijective.

所以你可以使用(未​​测试)

So you can use (not tested)

imul ax, ax, 0x6db7
cmp ax, 0x2492
ja not_multiple_of_7

这当然适用于无符号数.

This is for unsigned numbers of course.

作为奖励，ax 将是原始值除以 7 iff 它是 7 的倍数.

As a bonus, ax will be the original value divided by 7 iff it was a multiple of 7.

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