32 位处理器上的 64 位/64 位余数查找算法? [英] 64 bit / 64 bit remainder finding algorithm on a 32 bit processor?

问题描述

我知道过去曾有人问过类似的问题，但经过长时间的处理，我实现了使用重复减法除法正确找到商的算法.但是我无法从这种方法中找出剩余部分.有什么快速简便的方法可以在 32 位处理器上找出 64 位/64 位划分中的余数.更准确地说，我正在尝试实施

I know that similar questions has been asked in the past, but I have implemented after a long process the algorithm to find the quotient correctly using the division by repeated subtraction method. But I am not able to find out the remainder from this approach. Is there any quick and easy way for finding out remainder in 64bit/64bit division on 32bit processor. To be more precise I am trying to implement

ulldiv_t __aeabi_uldivmod(  
 unsigned long long n, unsigned long long d)  

在本文档中引用 http://infocenter.arm.com/help/topic/com.arm.doc.ihi0043d/IHI0043D_rtabi.pdf

推荐答案

什么?如果你做重复减法(这听起来很基础)，那么当你不能再做一次减法时，你剩下的就是余数不是那么简单吗?

What? If you do repeated subtraction (which sounds really basic), then isn't it as simple as whatever you have left when you can't do another subtraction is the remainder?

至少这是一种天真的直观方式:

At least that's the naïve intuitive way:

uint64_t simple_divmod(uint64_t n, uint64_t d)
{
  if (n == 0 || d == 0)
    return 0;
  uint64_t q = 0;
  while (n >= d)
  {
    ++q;
    n -= d;
  }
  return n;
}

或者我错过了这里的船?

Or am I missing the boat, here?

当然，这对于大数来说会非常慢，但这是重复减法.我敢肯定(即使不看！)还有更高级的算法.

Of course this will be fantastically slow for large numbers, but this is repeated subtraction. I'm sure (even without looking!) there are more advanced algorithms.

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