Convert int to a bit array in .NET

How can I convert an int to a bit array?

If I e.g. have an int with the value 3 I want an array, that has the length 8 and that looks like this:

0 0 0 0 0 0 1 1

Each of these numbers are in a separate slot in the array that have the size 8.

Use the BitArray class.

int value = 3;
BitArray b = new BitArray(new int[] { value });

If you want to get an array for the bits, you can use the BitArray.CopyTo method with a bool[] array.

bool[] bits = new bool[b.Count];
b.CopyTo(bits, 0);

Note that the bits will be stored from least significant to most significant, so you may wish to use Array.Reverse.

And finally, if you want get 0s and 1s for each bit instead of booleans (I'm using a byte to store each bit; less wasteful than an int):

byte[] bitValues = bits.Select(bit => (byte)(bit ? 1 : 0)).ToArray();

To convert the int 'x'

int x = 3;

One way, by manipulation on the int :

string s = Convert.ToString(x, 2); //Convert to binary in a string

int[] bits= s.PadLeft(8, '0') // Add 0's from left
	         .Select(c => int.Parse(c.ToString())) // convert each char to int
	         .ToArray(); // Convert IEnumerable from select to Array

Alternatively, by using the BitArray class-

BitArray b = new BitArray(new byte[] { x });
int[] bits = b.Cast<bool>().Select(bit => bit ? 1 : 0).ToArray();

Use Convert.ToString (value, 2)

so in your case

string binValue = Convert.ToString (3, 2);

I would achieve it in a one-liner as shown below:

using System;
using System.Collections;

namespace stackoverflowQuestions
{
    class Program
    {
        static void Main(string[] args)
        {    
            //get bit Array for number 20
            var myBitArray = new BitArray(BitConverter.GetBytes(20));
        }
    }
}

Please note that every element of a BitArray is stored as bool as shown in below snapshot:

So below code works:

if (myBitArray[0] == false)
{
    //this code block will execute
}

but below code doesn't compile at all:

if (myBitArray[0] == 0)
{
    //some code
}

I just ran into an instance where...

int val = 2097152;
var arr = Convert.ToString(val, 2).ToArray();
var myVal = arr[21];

...did not produce the results I was looking for. In 'myVal' above, the value stored in the array in position 21 was '0'. It should have been a '1'. I'm not sure why I received an inaccurate value for this and it baffled me until I found another way in C# to convert an INT to a bit array:

int val = 2097152;
var arr = new BitArray(BitConverter.GetBytes(val));
var myVal = arr[21];

This produced the result 'true' as a boolean value for 'myVal'.

I realize this may not be the most efficient way to obtain this value, but it was very straight forward, simple, and readable.

To convert your integer input to an array of bool of any size, just use LINQ.

bool[] ToBits(int input, int numberOfBits) {
	return Enumerable.Range(0, numberOfBits)
	.Select(bitIndex => 1 << bitIndex)
	.Select(bitMask => (input & bitMask) == bitMask)
	.ToArray();
}

So to convert an integer to a bool array of up to 32 bits, simply use it like so:

bool[] bits = ToBits(65, 8); // true, false, false, false, false, false, true, false

You may wish to reverse the array depending on your needs.

Array.Reverse(bits);

int value = 3;

var array = Convert.ToString(value, 2).PadLeft(8, '0').ToArray();

    public static bool[] Convert(int[] input, int length)
    {
        var ret = new bool[length];
        var siz = sizeof(int) * 8;
        var pow = 0;
        var cur = 0;

        for (var a = 0; a < input.Length && cur < length; ++a)
        {
            var inp = input[a];

            pow = 1;

            if (inp > 0)
            {
                for (var i = 0; i < siz && cur < length; ++i)
                {
                    ret[cur++] = (inp & pow) == pow;

                    pow *= 2;
                }
            }
            else
            {
                for (var i = 0; i < siz && cur < length; ++i)
                {
                    ret[cur++] = (inp & pow) != pow;

                    pow *= 2;
                }
            }
        }

        return ret;
    }

I recently discovered the C# Vector<T> class, which uses hardware acceleration (i.e. SIMD: Single-Instruction Multiple Data) to perform operations across the vector components as single instructions. In other words, it parallelizes array operations, to an extent.

Since you are trying to expand an integer bitmask to an array, perhaps you are trying to do something similar.

If you're at the point of unrolling your code, this would be an optimization to strongly consider. But also weigh this against the costs if you are only sparsely using them. And also consider the memory overhead, since Vectors really want to operate on contiguous memory (in CLR known as a Span<T>), so the kernel may be having to twiddle bits under the hood when you instantiate your own vectors from arrays.

---

Here is an example of how to do masking:

//given two vectors
Vector<int> data1 = new Vector<int>(new int[] { 1, 0, 1, 0, 1, 0, 1, 0 });
Vector<int> data2 = new Vector<int>(new int[] { 0, 1, 1, 0, 1, 0, 0, 1 });
//get the pairwise-matching elements
Vector<int> mask = Vector.Equals(data1, data2);
//and return values from another new vector for matches
Vector<int> whenMatched = new Vector<int>(new int[] { 1, 2, 3, 4, 5, 6, 7, 8 });
//and zero otherwise
Vector<int> whenUnmatched = Vector<int>.Zero;

//perform the filtering
Vector<int> result = Vector.ConditionalSelect(mask, whenMatched, whenUnmatched);
//note that only the first half of vector components render in the Debugger (this is a known bug)
string resultStr = string.Join("", result);
//resultStr is <0, 0, 3, 4, 5, 6, 0, 0>

^{Note that the VS Debugger is bugged, showing only the first half of the components of a vector.}

So with an integer as your mask, you might try:

int maskInt = 0x0F;//00001111 in binary
//convert int mask to a vector (anybody know a better way??)
Vector<int> maskVector = new Vector<int>(Enumerable.Range(0, Vector<int>.Count).Select(i => (maskInt & 1<<i) > 0 ? -1 : 0).ToArray());

Note that the (signed integer) -1 is used to signal true, which has binary representation of all ones.

Positive 1 does not work, and you can cast (int)-1 to uint to get every bit of the binary enabled, if needed (but not by using Enumerable.Cast<>()).

---

However this only works for int32 masks up to 2^8 because of the 8-element capacity in my system (that supports 4x64-bit chunks). This depends on the execution environment based on the hardware capabilities, so always use Vector<T>.Capacity.

You therefore can get double the capacity with shorts as ints as longs (the new Half type isn't yet supported, nor is "Decimal", which are the corresponding float/double types to int16 and int128):

ushort maskInt = 0b1111010101010101;
Vector<ushort> maskVector = new Vector<ushort>(Enumerable.Range(0, Vector<ushort>.Count).Select(i => (maskInt & 1<<i) > 0 ? -1 : 0).Select(x => (ushort)x).ToArray());
//string maskString = string.Join("", maskVector);//<65535, 0, 65535, 0, 65535, 0, 65535, 0, 65535, 0, 65535, 0, 65535, 65535, 65535, 65535>
Vector<ushort> whenMatched = new Vector<ushort>(Enumerable.Range(1, Vector<ushort>.Count).Select(i => (ushort)i).ToArray());//{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Vector<ushort> whenUnmatched = Vector<ushort>.Zero;

Vector<ushort> result = Vector.ConditionalSelect(maskVector, whenMatched, whenUnmatched);
string resultStr = string.Join("", result);//<1, 0, 3, 0, 5, 0, 7, 0, 9, 0, 11, 0, 13, 14, 15, 16>

---

Due to how integers work, being signed or unsigned (using the most significant bit to indicate +/- values), you might need to consider that too, for values like converting 0b1111111111111111 to a short. The compiler will usually stop you if you try to do something that appears to be stupid, at least.

short maskInt = unchecked((short)(0b1111111111111111));

Just make sure you don't confuse the most-significant bit of an int32 as being 2^31.

Using & (AND)

---

Bitwise Operators

The answers above are all correct and effective. If you wanted to do it old-school, without using BitArray or String.Convert(), you would use bitwise operators.

The bitwise AND operator & takes two operands and returns a value where every bit is either 1, if both operands have a 1 in that place, or a 0 in every other case. It's just like the logical AND (&&), but it takes integral operands instead of boolean operands.

Ex. 0101 & 1001 = 0001

Using this principle, any integer AND the maximum value of that integer is itself.

byte b = 0b_0100_1011; // In base 10, 75.
Console.WriteLine(b & byte.MaxValue); // byte.MaxValue = 255

Result: 75

---

Bitwise AND in a loop

We can use this to our advantage to only take specific bits from an integer by using a loop that goes through every bit in a positive 32-bit integer (i.e., uint) and puts the result of the AND operation into an array of strings which will all be "1" or "0".

A number that has a 1 at only one specific digit n is equal to 2 to the nth power (I typically use the Math.Pow() method).

public static string[] GetBits(uint x) {
    string[] bits = new string[32];
    
    for (int i = 0; i < 32; i++) {
        uint bit = x & Math.Pow(2, i);
        if (bit == 1)
            bits[i] = "1";
        else
            bits[i] = "0";
    }

    return bits;
}

If you were to input, say, 1000 (which is equivalent to binary 1111101000), you would get an array of 32 strings that would spell 0000 0000 0000 0000 0000 0011 1110 1000 (the spaces are just for readability).