是否有用于长类型的函数Math.Pow(A，n)?

本文介绍了是否有用于长类型的函数Math.Pow(A，n)?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

我正在测试一个小的C#程序片段:

I'm testing a small C# program fragment:

short min_short = (short)(int)-Math.Pow(2,15); short max_short = (short)(int)(Math.Pow(2, 15) - 1); Console.WriteLine("The min of short is:{0};\tThe max of short is:{1}", min_short, max_short); int min_int = (int)-Math.Pow(2, 31); int max_int = (int)(Math.Pow(2, 31) - 1); Console.WriteLine("The min of int is:{0};\tThe max of int is:{1}", min_int, max_int); uint min_uint = 0; uint max_uint = (uint)(Math.Pow(2, 32) - 1); Console.WriteLine("The min of uint is:{0};\tThe max of uint is:{1}", min_uint, max_uint); long min_long = (long)-Math.Pow(2, 63); long max_long = (long)(Math.Pow(2, 63) - 1); Console.WriteLine("The min of long is:{0};\tThe max of long is:{1}", min_long, max_long); ulong min_ulong = 0; ulong max_ulong = (ulong)(Math.Pow(2, 64) - 1); Console.WriteLine("The min of ulong is:{0};\tThe max of ulong is:{1}", min_ulong, max_ulong);

输出为:

The min of ushort is:0; The max of ushort is:65535 The min of short is:-32768; The max of short is:32767 The min of int is:-2147483648; The max of int is:2147483647 The min of uint is:0; The max of uint is:4294967295 The min of long is:-9223372036854775808;The max of long is:-9223372036854775808 The min of ulong is:0; The max of ulong is:0

我怀疑该错误是由Math.Pow()函数引起的，该函数返回的是双精度类型.

I suspect that the mistake is caused by the function of Math.Pow(),which is double type returned.

public static double Pow( double x, double y )

所以，我的问题是:长类型是否有类似的Math函数?如何更正上面程序片段中的错误.非常感谢！

so,my question is: Is there has a similar Math function for long type? How to correct the errors in the program fragment above. Many thanks!

推荐答案

您达到了 Math.Pow 的限制.您需要使用 System.Numerics.BigInteger.Pow .

You reached Math.Pow limit. You would need to use System.Numerics.BigInteger.Pow.