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Replacing bits in a Byte Array C#

I want to open a Bitmap File in C# as an array of bytes, and replace certain bytes within that array, and rewrite the Byte array back to disk as a bitmap again.
My current approach is to read into a byte[] array, then convert that array to a list to begin editing individual bytes.
originalBytes = File.ReadAllBytes(path);
List<byte> listBytes = new List<Byte>(originalBytes);
How does one go about replacing every nth byte in the array with a user configured/different byte each time and rewriting back to file?

no need in List<byte>
replaces every n-th byte with customByte
var n = 5;
byte customByte = 0xFF;
var bytes = File.ReadAllBytes(path);
for (var i = 0; i < bytes.Length; i++)
{
if (i%n == 0)
{
bytes[i] = customByte;
}
}
File.WriteAllBytes(path, bytes);

Assuming that you want to replace every nth byte with the same new byte, you could do something like this (shown for every 3rd byte):
int n = 3;
byte newValue = 0xFF;
for (int i = n; i < listBytes.Count; i += n)
{
listBytes[i] = newValue;
}
File.WriteAllBytes(path, listBytes.ToArray());
Of course, you could also do this with a fancy LINQ expression which would be harder to read i guess.

Technically, you can implement something like this:
// ReadAllBytes returns byte[] array, we have no need in List<byte>
byte[] data = File.ReadAllBytes(path);
// starting from 0 - int i = 0 - will ruin BMP header which we must spare
// if n is small, you may want to start from 2 * n, 3 * n etc.
// or from some fixed offset
for (int i = n; i < data.Length; i += n)
data[i] = yourValue;
File.WriteAllBytes(path, data);
Please notice, that Bitmap file has a header
https://en.wikipedia.org/wiki/BMP_file_format
that's why I've started loop from n, not from 0

Get char array from byte array and then back to byte array

I don't know why but when you do the next thing you will never get the same as the original byte array:
var b = new byte[] {252, 2, 56, 8, 9};
var g = System.Text.Encoding.ASCII.GetChars(b);
var f = System.Text.Encoding.ASCII.GetBytes(g);
If you will run this code you will see that b != f, Why?!
Is there any way to convert bytes to chars and then back to bytes and get the same as the original byte array?

byte value can be 0 to 255.
When the byte value > 127, then result of
System.Text.Encoding.ASCII.GetChars()
is always '?' which has value 63
Therefore,
System.Text.Encoding.ASCII.GetBytes()
result always get 63 (wrong value) for those have initial byte value > 127
If you need TABLE ASCII -II then you can do as following
var b = new byte[] { 252, 2, 56, 8, 9 };
//another encoding
var e = Encoding.GetEncoding("437");
//252 inside the mentioned table is ⁿ and now you have it
var g = e.GetString(b);
//now you can get the byte value 252
var f = e.GetBytes(g);
Similar posts you can read
How to convert the byte 255 to a signed char in C#
How can I convert extended ascii to a System.String?

Why not use chars?
var b = new byte[] {252, 2, 56, 8, 9};
var g = new char[b.Length];
var f = new byte[g.Length]; // can also be b.Length, doens't really matter
for (int i = 0; i < b.Length; i++)
{
g[i] = Convert.ToChar(b[i]);
}
for (int i = 0; i < f.Length; i++)
{
f[i] = Convert.ToByte(g[i]);
}

The only difference is first byte: 252. Because ascii char is 1-byte signed char and it's value range is -128 to 127. Actually your input is incorrect. signed char can't be 252.

Converting 3 bytes into signed integer in C#

I'm trying to convert 3 bytes to signed integer (Big-endian) in C#.
I've tried to use BitConverter.ToInt32 method, but my problem is what value should have the lats byte.
Can anybody suggest me how can I do it in different way?
I also need to convert 5 (or 6 or 7) bytes to signed long, is there any general rule how to do it?
Thanks in advance for any help.

As a last resort you could always shift+add yourself:
byte b1, b2, b3;
int r = b1 << 16 | b2 << 8 | b3;
Just swap b1/b2/b3 until you have the desired result.
On second thought, this will never produce negative values.
What result do you want when the msb >= 0x80 ?
Part 2, brute force sign extension:
private static int Bytes2Int(byte b1, byte b2, byte b3)
{
int r = 0;
byte b0 = 0xff;
if ((b1 & 0x80) != 0) r |= b0 << 24;
r |= b1 << 16;
r |= b2 << 8;
r |= b3;
return r;
}
I've tested this with:
byte[] bytes = BitConverter.GetBytes(p);
int r = Bytes2Int(bytes[2], bytes[1], bytes[0]);
Console.WriteLine("{0} == {1}", p, r);
for several p.

The last value should be 0 if it isn't set for a positive number, 256 for a negative.
To know what you should pass in, you can try converting it the other way:
var bytes = BitConverter.GetBytes(i);
int x = BitConverter.ToInt32(bytes, 0);

To add to the existing answers here, there's a bit of a gotcha in that Bitconverter.ToInt32() will throw an ArgumentException if the array is less than sizseof(int) (4) bytes in size;
Destination array is not long enough to copy all the items in the collection. Check array index and length.
Given an array less than sizeof(int) (4) bytes in size, you can compensate for left/right padding like so;
Right-pad
Results in positive Int32 numbers
int intByteSize = sizeof(int);
byte[] padded = new byte[intByteSize];
Array.Copy(sourceBytes, 0, padded, 0, sourceBytes.Length);
sourceBytes = padded;
Left-pad
Results in negative Int32 numbers, assuming non-zero value at byte index sourceBytes.Length - 1.
int intByteSize = sizeof(int);
byte[] padded = new byte[intByteSize];
Array.Copy(sourceBytes, 0, padded, intByteSize - sourceBytes.Length, sourceBytes.Length);
sourceBytes = padded;
Once padded, you can safely call int myValue = BitConverter.ToInt32(sourceBytes, 0);.

Convert Byte Array to Bit Array?

How would I go about converting a bytearray to a bit array?

The obvious way; using the constructor that takes a byte array:
BitArray bits = new BitArray(arrayOfBytes);

It depends on what you mean by "bit array"... If you mean an instance of the BitArray class, Guffa's answer should work fine.
If you actually want an array of bits, in the form of a bool[] for instance, you could do something like that :
byte[] bytes = ...
bool[] bits = bytes.SelectMany(GetBits).ToArray();
...
IEnumerable<bool> GetBits(byte b)
{
for(int i = 0; i < 8; i++)
{
yield return (b & 0x80) != 0;
b *= 2;
}
}

public static byte[] ToByteArray(this BitArray bits)
{
int numBytes = bits.Count / 8;
if (bits.Count % 8 != 0) numBytes++;
byte[] bytes = new byte[numBytes];
int byteIndex = 0, bitIndex = 0;
for (int i = 0; i < bits.Count; i++) {
if (bits[i])
bytes[byteIndex] |= (byte)(1 << (7 - bitIndex));
bitIndex++;
if (bitIndex == 8) {
bitIndex = 0;
byteIndex++;
}
}
return bytes;
}

You can use BitArray to create a stream of bits from a byte array. Here an example:
string testMessage = "This is a test message";
byte[] messageBytes = Encoding.ASCII.GetBytes(testMessage);
BitArray messageBits = new BitArray(messageBytes);

byte number = 128;
Convert.ToString(number, 2);
=> out: 10000000

public static byte[] ToByteArray(bool[] byteArray)
{
return = byteArray
.Select(
(val1, idx1) => new { Index = idx1 / 8, Val = (byte)(val1 ? Math.Pow(2, idx1 % 8) : 0) }
)
.GroupBy(gb => gb.Index)
.Select(val2 => (byte)val2.Sum(s => (byte)s.Val))
.ToArray();
}

Why do I get the following output when inverting bits in a byte?

Assumption:
Converting a
byte[] from Little Endian to Big
Endian means inverting the order of the bits in
each byte of the byte[].
Assuming this is correct, I tried the following to understand this:
byte[] data = new byte[] { 1, 2, 3, 4, 5, 15, 24 };
byte[] inverted = ToBig(data);
var little = new BitArray(data);
var big = new BitArray(inverted);
int i = 1;
foreach (bool b in little)
{
Console.Write(b ? "1" : "0");
if (i == 8)
{
i = 0;
Console.Write(" ");
}
i++;
}
Console.WriteLine();
i = 1;
foreach (bool b in big)
{
Console.Write(b ? "1" : "0");
if (i == 8)
{
i = 0;
Console.Write(" ");
}
i++;
}
Console.WriteLine();
Console.WriteLine(BitConverter.ToString(data));
Console.WriteLine(BitConverter.ToString(ToBig(data)));
foreach (byte b in data)
{
Console.Write("{0} ", b);
}
Console.WriteLine();
foreach (byte b in inverted)
{
Console.Write("{0} ", b);
}
The convert method:
private static byte[] ToBig(byte[] data)
{
byte[] inverted = new byte[data.Length];
for (int i = 0; i < data.Length; i++)
{
var bits = new BitArray(new byte[] { data[i] });
var invertedBits = new BitArray(bits.Count);
int x = 0;
for (int p = bits.Count - 1; p >= 0; p--)
{
invertedBits[x] = bits[p];
x++;
}
invertedBits.CopyTo(inverted, i);
}
return inverted;
}
The output of this little application is different from what I expected:
00000001 00000010 00000011 00000100 00000101 00001111 00011000
00000001 00000010 00000011 00000100 00000101 00001111 00011000
80-40-C0-20-A0-F0-18
01-02-03-04-05-0F-18
1 2 3 4 5 15 24
1 2 3 4 5 15 24
For some reason the data remains the same, unless printed using BitConverter.
What am I not understanding?
Update
New code produces the following output:
10000000 01000000 11000000 00100000 10100000 11110000 00011000
00000001 00000010 00000011 00000100 00000101 00001111 00011000
01-02-03-04-05-0F-18
80-40-C0-20-A0-F0-18
1 2 3 4 5 15 24
128 64 192 32 160 240 24
But as I have been told now, my method is incorrect anyway because I should invert the bytes
and not the bits?
This hardware developer I'm working with told me to invert the bits because he cannot read the data.
Context where I'm using this
The application that will use this does not really work with numbers.
I'm supposed to save a stream of bits to file where
1 = white and 0 = black.
They represent pixels of a bitmap 256x64.
byte 0 to byte 31 represents the first row of pixels
byte 32 to byte 63 the second row of pixels.
I have code that outputs these bits... but the developer is telling
me they are in the wrong order... He says the bytes are fine but the bits are not.
So I'm left confused :p

No. Endianness refers to the order of bytes, not bits. Big endian systems store the most-significant byte first and little-endian systems store the least-significant first. The bits within a byte remain in the same order.
Your ToBig() function is returning the original data rather than the bit-swapped data, it seems.

Your method may be correct at this point. There are different meanings of endianness, and it depends on the hardware.
Typically, it's used for converting between computing platforms. Most CPU vendors (now) use the same bit ordering, but different byte ordering, for different chipsets. This means, that, if you are passing a 2-byte int from one system to another, you leave the bits alone, but swap bytes 1 and 2, ie:
int somenumber -> byte[2]: somenumber[high],somenumber[low] ->
byte[2]: somenumber[low],somenumber[high] -> int newNumber
However, this isn't always true. Some hardware still uses inverted BIT ordering, so what you have may be correct. You'll need to either trust your hardware dev. or look into it further.
I recommend reading up on this on Wikipedia - always a great source of info:
http://en.wikipedia.org/wiki/Endianness
Your ToBig method has a bug.
At the end:
invertedBits.CopyTo(data, i);
}
return data;
You need to change that to:
byte[] newData = new byte[data.Length];
invertedBits.CopyTo(newData, i);
}
return newData;
You're resetting your input data, so you're receiving both arrays inverted. The problem is that arrays are reference types, so you can modify the original data.

As greyfade already said, endianness is not about bit ordering.
The reason that your code doesn't do what you expect, is that the ToBig method changes the array that you send to it. That means that after calling the method the array is inverted, and data and inverted are just two references pointing to the same array.
Here's a corrected version of the method.
private static byte[] ToBig(byte[] data) {
byte[] result = new byte[data.length];
for (int i = 0; i < data.Length; i++) {
var bits = new BitArray(new byte[] { data[i] });
var invertedBits = new BitArray(bits.Count);
int x = 0;
for (int p = bits.Count - 1; p >= 0; p--) {
invertedBits[x] = bits[p];
x++;
}
invertedBits.CopyTo(result, i);
}
return result;
}
Edit:
Here's a method that changes endianness for a byte array:
static byte[] ConvertEndianness(byte[] data, int wordSize) {
if (data.Length % wordSize != 0) throw new ArgumentException("The data length does not divide into an even number of words.");
byte[] result = new byte[data.Length];
int offset = wordSize - 1;
for (int i = 0; i < data.Length; i++) {
result[i + offset] = data[i];
offset -= 2;
if (offset < -wordSize) {
offset += wordSize * 2;
}
}
return result;
}
Example:
byte[] data = { 1,2,3,4,5,6 };
byte[] inverted = ConvertEndianness(data, 2);
Console.WriteLine(BitConverter.ToString(inverted));
Output:
02-01-04-03-06-05
The second parameter is the word size. As endianness is the ordering of bytes in a word, you have to specify how large the words are.
Edit 2:
Here is a more efficient method for reversing the bits:
static byte[] ReverseBits(byte[] data) {
byte[] result = new byte[data.Length];
for (int i = 0; i < data.Length; i++) {
int b = data[i];
int r = 0;
for (int j = 0; j < 8; j++) {
r <<= 1;
r |= b & 1;
b >>= 1;
}
result[i] = (byte)r;
}
return result;
}

One big problem I see is ToBig changes the contents of the data[] array that is passed to it.
You're calling ToBig on an array named data, then assigning the result to inverted, but since you didn't create a new array inside ToBig, you modified both arrays, then you proceed to treat the arrays data and inverted as different when in reality they are not.