我正在尋找一種有效的方法來確定設置在整數中的最低有效位的位置，例如對于 0x0FF0，它將是 4.
一個簡單的實現是這樣的:
unsigned GetLowestBitPos(無符號值){斷言(值！= 0)；//單獨處理無符號位置 = 0;while (!(value & 1)){值＞＞＝1；++位置；}返回位置；}任何想法如何從中擠出一些周期?
(注意:這個問題是針對喜歡這種東西的人，而不是告訴我xyz優化是邪惡的.)
[edit] 感謝大家的想法！我也學到了一些其他的東西.酷！
Bit Twiddling Hacks
a> 提供了一個很好的集合，呃，一些小技巧，并附有性能/優化討論.對于您的問題(來自該站點)，我最喜歡的解決方案是乘法和查找":
unsigned int v;//在 32 位 v 中找到尾隨零的數量國際r;//結果在這里static const int MultiplyDeBruijnBitPosition[32] ={0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9};r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >>27];有用的參考資料:
I am looking for an efficient way to determine the position of the least significant bit that is set in an integer, e.g. for 0x0FF0 it would be 4.
A trivial implementation is this:
unsigned GetLowestBitPos(unsigned value)
{
assert(value != 0); // handled separately
unsigned pos = 0;
while (!(value & 1))
{
value >>= 1;
++pos;
}
return pos;
}
Any ideas how to squeeze some cycles out of it?
(Note: this question is for people that enjoy such things, not for people to tell me xyzoptimization is evil.)
[edit] Thanks everyone for the ideas! I've learnt a few other things, too. Cool!
Bit Twiddling Hacks offers an excellent collection of, er, bit twiddling hacks, with performance/optimisation discussion attached. My favourite solution for your problem (from that site) is ?multiply and lookup?:
unsigned int v; // find the number of trailing zeros in 32-bit v
int r; // result goes here
static const int MultiplyDeBruijnBitPosition[32] =
{
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >> 27];
Helpful references:
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