Initialize BitArray from integer

Question

I'm trying to initialize a System.BitArray instance from integer value. However, it looks like I don't get the right values.

My code is

        var b = new BitArray(BitConverter.GetBytes(0xfa2));
        for (int i = 0; i < b.Count; i++)
        {
            char c = b[i] ? '1' : '0';
            Console.Write(c);
        }
        Console.WriteLine();

I've tried also without BitConverter:

        var b = new BitArray(new int[] { 0xfa2 });

But none of these attempt seem to work. These are the attempts that was suggested here: Convert int to a bit array in .NET

My output: 01000101111100000000000000000000. The excpected output: 111110100010.

Any help will be really appreciated!

It has to do with the endienness of the architecture. Possible duplicate of http://stackoverflow.com/questions/217980/c-sharp-little-endian-or-big-endian — Wagner DosAnjos, Aug 29 '14 at 17:57
do you expect 12 bits instead of 32? or are you talking about the order of the bits? — RadioSpace, Aug 29 '14 at 18:09

score 5 · Accepted Answer · answered Aug 29 '14 at 17:56

5

You are looping from the wrong direction. Try this:

    var b = new BitArray(BitConverter.GetBytes(0xfa2));
    for (int i = b.Count-1; i >= 0; i--)
    {
        char c = b[i] ? '1' : '0';
        Console.Write(c);
    }
    Console.WriteLine();

answered Aug 29 '14 at 17:56

vcsjones

138,677
31
291
286

Wagner DosAnjos · Answer 2 · 2014-08-29T18:29:40.943

This is a Little-Endian vs Big-Endian issue, so you need to take the endienness of the hw architecture into consideration. Also based on the documentation you need to change the way you print the BitArray.

byte[] buffer = BitConverter.GetBytes((ushort)0xfa2);

if (BitConverter.IsLittleEndian) Array.Reverse(buffer);

var b = new BitArray(buffer);

for (int i = 0; i < b.Count; i+=8)
{
    for (int j=i + 7; j >= i; j--)
    {
        char c = b[j] ? '1' : '0';
        Console.Write(c);
    }
}

Console.WriteLine();

Reference:

From BitArray Constructor documentation

The first byte in the array represents bits 0 through 7, the second byte represents bits 8 through 15, and so on. The Least Significant Bit of each byte represents the lowest index value: " bytes [0] & 1" represents bit 0, " bytes [0] & 2" represents bit 1, " bytes [0] & 4" represents bit 2, and so on.

Rimas · Answer 3 · 2014-08-29T18:52:52.120

As stated in documentation BitArray Constructor (Int32[]):

The number in the first values array element represents bits 0 through 31, the second number in the array represents bits 32 through 63, and so on. The Least Significant Bit of each integer represents the lowest index value: " values [0] & 1" represents bit 0, " values [0] & 2" represents bit 1, " values [0] & 4" represents bit 2, and so on.

When use this constructor there no need to check endianness, just reverse output order:

var b = new BitArray(new int[] { 0xfa2 });
// skip leading zeros, but leave least significant bit:
int count = b.Count;
while (count > 1 && !b[count-1])
    count--;
// output
for (int i = count - 1; i >= 0; i--)
{
    char c = b[i] ? '1' : '0';
    Console.Write(c);
}
Console.WriteLine();

Initialize BitArray from integer

3 Answers3