与数学幂加数相反的方法 [英] Opposite method of math power adding numbers

问题描述

我有将布尔数组转换为整数的方法.看起来像这样

I have method for converting array of Booleans to integer. It looks like this

class Program
    {
        public static int GivMeInt(bool[] outputs)
        {
            int data = 0;
            for (int i = 0; i < 8; i++)
            {
                data += ((outputs[i] == true) ? Convert.ToInt32(Math.Pow(2, i)) : 0);
            }
            return data;

        }

        static void Main(string[] args)
        {
            bool[] outputs = new bool[8];
            outputs[0] = false;
            outputs[1] = true;
            outputs[2] = false;
            outputs[3] = true;
            outputs[4] = false;
            outputs[5] = false;
            outputs[6] = false;
            outputs[7] = false;
            int data = GivMeInt(outputs);
            Console.WriteLine(data);
            Console.ReadKey();
        }
   }

现在，我想使用相反的方法返回布尔值数组 到目前为止，由于我对.NET和C#的了解不足，所以我只对switch语句或每个可能的int值的条件进行了硬编码.

Now I want to make opposite method returning array of Booleans values As I am short with knowledge of .NET and C# until now I have only my mind hardcoding of switch statement or if conditions for every possible int value.

public static bool[] GiveMeBool(int data)
        {
            bool[] outputs = new bool[8];

            if (data == 0)
            {
                outputs[0] = false;
                outputs[1] = false;
                outputs[2] = false;
                outputs[3] = false;
                outputs[4] = false;
                outputs[5] = false;
                outputs[6] = false;
                outputs[7] = false;
            }
            //After thousand lines of coed
            if (data == 255)
            {
                outputs[0] = true;
                outputs[1] = true;
                outputs[2] = true;
                outputs[3] = true;
                outputs[4] = true;
                outputs[5] = true;
                outputs[6] = true;
                outputs[7] = true;
            }
            return outputs;
        }

我知道必须有更简单的方法.

I know that there must be easier way.

推荐答案

您需要使用按位运算符:(已测试)

You need to use bitwise operators: (Tested)

public static bool[] GiveMeBool(int data) {
    bool[] outputs = new bool[8];

    for(int i = 0; i < 8; i++)
        outputs[i] = (data & (1 << i)) == (1 << i);

    return outputs;
}

您还可以使用按位运算符使原始的很多更快:(未经测试)

You can also use bitwise operators to make your original much faster: (Untested)

public static int GivMeInt(bool[] outputs) {
    int data = 0;
    for (int i = 0; i < 8; i++) 
        data += outputs[i] ? 1 << i : 0;

    return data;
}

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