问题描述
.NET 中的内置 Math.Pow() 函数将 double 基数提升为 double 指数并返回 >double 结果.
The built-in Math.Pow() function in .NET raises a double base to a double exponent and returns a double result.
用整数做同样的事情的最好方法是什么?
What's the best way to do the same with integers?
补充:似乎可以将 Math.Pow() 结果转换为 (int)，但这总是会产生正确的数字且不会出现舍入错误吗?
Added: It seems that one can just cast Math.Pow() result to (int), but will this always produce the correct number and no rounding errors?
推荐答案
一个相当快的可能是这样的:
A pretty fast one might be something like this:
int IntPow(int x, uint pow)
{
int ret = 1;
while ( pow != 0 )
{
if ( (pow & 1) == 1 )
ret *= x;
x *= x;
pow >>= 1;
}
return ret;
}
请注意，这不允许负幂.我会把它作为练习留给你.:)
Note that this does not allow negative powers. I'll leave that as an exercise to you. :)
已添加:哦，是的，差点忘了 - 还要添加上溢/下溢检查，否则您可能会遇到一些令人讨厌的惊喜.
Added: Oh yes, almost forgot - also add overflow/underflow checking, or you might be in for a few nasty surprises down the road.
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