有什么方法可以将64位整数除以10的RARS？

发布于 2025-02-02 18:25:37 字数 602 浏览 5 评论 0原文

我一直在尝试将两个32位登记册除以10。我知道如果股息为32位，我知道如何分配寄存器，但是如果它是64位。
我无法解决

div     s8, s0, a6 # divide a result by 10 (for example, 1234->123)
mul     s6, s8, a6 # multiply s8 by 10 (for example, 123->1230)
sub     s7, s0, s6 # the result - s6 (for example, 1234-1230->4)
beqz    s7, output # if s7 is zero, jump

。

该部门更棘手。您可以通过comapare，移位和减法实现常规二进制划分，也可以使用另一种方法：考虑到A * 2^32 + b，请考虑64位号AB（A和BE为32位数字）。表示2 ^32为x * 10 + y。然后ab / 10 = x +（y *a + b） / 10。如果您两次使用此方法，则可以将64位的划分降低10至32位。< / p>

但是我不明白，如果我两次使用此方法，为什么我可以获得32位号。

原文

I have been trying to divide two 32-bit registers by 10.
I know how to divide a register if a dividend is 32-bit like the following code, but I cannot solve if it's 64-bit.

div     s8, s0, a6 # divide a result by 10 (for example, 1234->123)
mul     s6, s8, a6 # multiply s8 by 10 (for example, 123->1230)
sub     s7, s0, s6 # the result - s6 (for example, 1234-1230->4)
beqz    s7, output # if s7 is zero, jump

And I found a tips that

The division is more tricky. You may either implement regular binary division by comapare, shift and subtract or use another method: Think of a 64bit number ab (where a and be are 32-bit numbers) as of a * 2^32 + b. Express 2 ^32 as x * 10 + y. Then ab / 10 = x + (y *a + b) / 10. If you apply this method twice, you may reduce 64-bit division by 10 to a 32-bit one.

But I cannot understand why I can get a 32-bit number if I apply this method twice.

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