Represent a list/array of integers in a 64 bit array

Question

0.00/5 (No votes)

See more:

I have a List/array with some integers.
array = {1,2,3,4,5,7,11,14,15...}
for each integer that exists in the array/list from 1 to 8 i append a bit until 1 byte and then convert the binary to decimal. In this case 11111010(6 and 8 does not exist) = 250. Likewise the next iteration.

What I have tried:

I tried it with appending each bit to a string when the number occurs and then converting it to decimal. Once all 8 iterations done, convert it into an 8 byte array.
Is there a better way to do it? perhaps using bitwise operators?

C#

private byte[] GetBitMapFromFields(List<int> fieldNo)
        {
            byte[] bytes = new byte[8];
            List<int> tempList = new List<int>();
            string binaryfield = string.Empty;
            var count = 1;
            for (var i = 0; i < fieldNo.Count; i++)
            {
                var k = 0;
                for(var j =1; j<=8;j++)
                {
                    if (count == fieldNo[i])
                    {
                        binaryfield = binaryfield + 1;
                        i++;
                        k++;
                        count++;
                    }
                    else
                    {
                        binaryfield = binaryfield + 0;
                        k++;
                        count++;
                    }
                    if(k==8)
                    {
                        i--;
                        break;
                    }
                }
                int output = Convert.ToInt32(binaryfield, 2);
                binaryfield = string.Empty;
                tempList.Add(output);

            }
            bytes = ConvertListToByte(tempList); //Converts the list to byte array of length 8
            return bytes;
        }

Posted 14-May-20 22:26pm

Jacxx

Updated 16-May-20 11:40am

Add a Solution

3 solutions

Solution 1

Something like the following will capture eight bits at a time:

C++

int pattern = 0;

for ( // expression to iterate over your list
{
    if ( // if the value exists
    {
        pattern |= 1;
    }
    pattern <<= 1;
}

// pattern now contains the mask of the items.

You also need to break the loop after 8 test items. Alternatively you could capture 32 or 64 bits at a time.

Posted 14-May-20 22:49pm

Richard MacCutchan

Solution 2

I did something similar when encrypting data using the BitArray class. Hope this is of some help

C#

static void Main(string[] args)
       {
           List<int> numbers = new List<int> { 1, 2, 3, 4, 5, 7 };//, 11, 14, 15
           int max = numbers.Max();
           int mod8 = max % 8;
           int length = mod8 == 0 ? max : max + 8 - mod8;
           BitArray bits = new BitArray(length);
           foreach (var n in numbers) bits[length - n] = true;
           int[] tempIntArray = new int[1];
           bits.CopyTo(tempIntArray, 0);
           int result = tempIntArray[0];

           Console.ReadLine();
       }

Posted 16-May-20 5:27am

George Swan

Comments

Jacxx 17-May-20 8:33am

Thank you George, that solution did work. However, instead of int[] i converted the bits directly to byte[] -
byte[] byteArray = new byte[(int)Math.Ceiling((double)bits.Length / 8)];
bits.CopyTo(byteArray, 0);
Also had to reverse it.

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Patrice T · Accepted Answer · 2020-05-16T11:41:00

Your code to set bits look rather complicated.
First of all you need to know that an integer is stored in base 2, the integer is a field of bits.
example of 10!

C#

10= 1*2^3+0*2^2+1*2^1+0*2^0

bits are numbered from right to left, starting with bit 0 and number match the powers of 2 from previous formula.

C#

bit numbers: 7 6 5 4 3 2 1 0
Values:                    ^ = 1 = 2^0
                         ^ = 2 = 2^1
                       ^ = 4 = 2^2
                      ^ = 8 = 2^3
...
              ^ = 128 = 2^7
10 is         0 0 0 0 1 0 1 0

To set high bits in the 8 bits, you just have to use the knowledge of their values and know bitwise operations.
a single loop does the trick:

C#

int set8bits (int value) { // set high bits
    int bit= 128;
    for ( int i=7, i>=0, i--) {
        if (value & bit) { // is true when bit is already set
            break;
        }
        value = value | bit; // set bit
        bit= bit >> 1 // move next bit
    }
    return value;
}