Question

I m trying to avoid long longs and integer overflow in some calculations, so I came up with the function below to calculate (a * b) / c (order is important due to truncating integer division).

 unsigned muldiv(unsigned a, unsigned b, unsigned c)
 {
      return a * (b / c) + (a * (b % c)) / c;
 }

是否有任何最优的案例,这些案例会像预期的那样赢得工作?

Answer 1

EDITED: This is correct for a superset of values for which the original obvious logic was correct. It still buys you nothing if c > b and possibly in other conditions. Perhaps you know something about your values of c but this may not help as much as you expect. Some combinations of a, b, c will still overflow.

EDIT:由于严格的C++98便携式原因,假设你重新避免<条码>长期,你可以通过促进<条码>至<条码>的<>unsign至<条码>,从而获得大约52个精度参数,这些参数很可能具有完成数学的整体价值。事实上,使用<条码>杜布尔的数学可能比三个整体司要快。

Answer 2

这在几个案件中是失败的。最为明显的是,<代码>a是大型的,因此a * (b % c) 超支。在此情况下,你可以尝试打字<条码>a和<条码>b,但如<条码>、<条码>、<条码>、<条码>和<条码>仍为之。审议a =b = 2^25-1 and c = 2^24 with a 32 bit unsign. 正确结果为2^26-4,但a * (b % c)和b* (a % c)均将超支。甚至<代码>(c %) * (b % c)也会流出。

迄今为止,解决这一问题的总体最突出的途径是扩大倍数,使中间产品更加精确。如果你没有这样做的话,你需要将其综合起来,摆脱较小的多层次和分裂,这在很大程度上与执行你自己的大面积图书馆相同。

如果你可以认为<条码>c小到<条码>(c-1)*(c-1)不会溢出未签名,你可以使用:

unsigned muldiv(unsigned a, unsigned b, unsigned c) {
    return (a/c)*(b/c)*c + (a%c)*(b/c) + (a/c)*(b%c) + (a%c)*(b%c)/c;
}

这实际上将给你所有a和b的“正确”答案——a * b)/c %(UINT_max+1)

Answer 3

To avoid overflow you have to pre-divide and then post-multiply by some factor.

最佳使用系数为c,只要(或两者)大于c。这就是Chris Dodd的职能。它的中间体最大((c % c) * (b % c)),正如Chris指出的,中间体小于或等于(c-1)*(c-1)。

If you could have a situation where both a and b are less than c, but (a * b) could still overflow, (which might be the case when c approaches the limit of the word size) then the best factor to use is a large power of two, to turn the multiply and divides into shifts. Try shifting by half the word size.

请注意,如果你没有更长的字句,则使用预告器,然后使用多面字。假设你不要抛弃低秩序轨道,而只是将其作为另一个术语,那么你只是使用几个字而不是一个大字。

我恳请你填写该守则。

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