The opposite method of adding mathematical energy

Question

I have a method to convert an array of Booleans to an integer. 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();
        }
   }

Now I want to make the opposite method, returning an array of Booleans values. Since I have been briefly familiar with .NET and C # so far, I only have my mental coding of the switch statement or condition for each 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;
        }

I know there should be an easier way.

Answer 1

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;
}

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;
}

Answer 2

This uses a bit offset.

public static bool[] GiveMeBool(Int32 data)
{
    bool[] bits = new bool[32];

    for (i = 0; i <= bits.Length - 1; i++) {
        bits(i) = (data & 1) == 1;
        data >>= 1;
    }

    return bits;
}

Answer 3

All this can be changed to be used with the help of shift operators .

Answer 4

Try it...

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

    for (int i = 0; i < 8; i++)
        outputs[i] = (data & (int)Math.Pow(2, i)) != 0;

    return outputs;
}

Answer 5

How about the following:

Int32 x = 0xFF33;
bool[] retval = new bool[32];
for (int i = 0; i < 32 && x != 0; i++, x = x >> 1)
{
    retval[i] = (x & 1) == 1;
}

He uses a beat shift to do his magic.

Answer 6

I tested the methods in Java. The only significant difference is the boolvs keywords boolean.

public class Test {

    public static void main(String[] args) {
        boolean[] bools = new boolean[]{true,true,false,false,false,false,false,false};
        int num = GivMeInt(bools);
        boolean[] bools2 = GivMeBools(num);
        for(int i = 0; i < 8; i++) System.out.println(bools[i]==bools2[i]);
    }

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

                }
            }
            return data;
    }

    public static boolean[] GivMeBools(int input)
        {
            boolean[] outputs = new boolean[8];
            for (int i = 0; i < 8; i++)
            {
                outputs[i] = (input & 0x1) == 1;
                input = input >>> 1;
            }
            return outputs;

        }
}