问题描述
我需要找到大于或等于给定值的 2 的最小幂.到目前为止，我有这个:
I need to find the smallest power of two that's greater or equal to a given value. So far, I have this:
int value = 3221; // 3221 is just an example, could be any number
int result = 1;
while (result < value) result <<= 1;
它工作正常，但感觉有点天真.有没有更好的算法来解决这个问题?
It works fine, but feels kind of naive. Is there a better algorithm for that problem?
编辑.有一些不错的汇编程序建议，所以我将这些标签添加到问题中.
EDIT. There were some nice Assembler suggestions, so I'm adding those tags to the question.
推荐答案
这是我最喜欢的.除了对它是否无效的初始检查(<0，如果你知道你只传入了 >=0 的数字，你可以跳过它)，它没有循环或条件，因此将优于大多数其他方法.这与 erickson 的答案类似，但我认为我在开头递减 x 并在结尾加 1 比他的答案少一些尴尬(并且也避免了结尾的条件).
Here's my favorite. Other than the initial check for whether it's invalid (<0, which you could skip if you knew you'd only have >=0 numbers passed in), it has no loops or conditionals, and thus will outperform most other methods. This is similar to erickson's answer, but I think that my decrementing x at the beginning and adding 1 at the end is a little less awkward than his answer (and also avoids the conditional at the end).
/// Round up to next higher power of 2 (return x if it's already a power
/// of 2).
inline int
pow2roundup (int x)
{
if (x < 0)
return 0;
--x;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x+1;
}
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