本文介绍了你如何在 C# 中做 *integer* 幂运算?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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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07-31 19:30