是否有用于长类型的函数Math.Pow(A，n)? [英] Is there has a function Math.Pow(A,n) for long type?

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.

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