Develop Reference

c# how to store two 4 bit numbers into one byte

So I have two 4 bits number (between 0 and 15):
int a = 1;
int b = 15;
And I want to store them into 1 byte.
This is what I have try:
byte[] bytesA = BitConverter.GetBytes(a);
byte[] bytesB = BitConverter.GetBytes(b);
byte a = bytesA[0];
byte b = bytesB[1];

Sins a byte is 8 bits and you have 2 4 bit values, you will need to shift one of the values to occupy the high 4 bits of the resulting byte, like so:
byte ab = (byte)((a << 4) | (b & 0xF));
To get a out of ab:
byte a = (byte)((ab >> 4) & 0xF);
To get b out of ab:
byte b = (byte)(ab & 0xF);

How to perform bit copy in C# or byte array left shift operation

I want to store information in concatenated fashion in a byte array (2 Bits + 6 Byte + 14 Bits) but I have no idea how to do it. With Buffer.Blockcopy, I can work with bytes, not bits. Moreover, 6 Byte data is already a part of data structure and referenced everywhere, so I don't wish to change that variable. I was thinking of declaring a byte then making a binary string cutting and concatenating, but I think it will be a very bad implementation. There must be a better and efficient way of doing this.
Here is the C# script that I am currently using, but I think there must be better and efficient way to do this.
byte[] ARC = new byte[7]{0x00,0xB1,0x1C,0x3D,0x4C,0x1A,0xEF};
RotateLeft(ARC);
RotateLeft(ARC);
RotateLeft(ARC);
RotateLeft(ARC);
RotateLeft(ARC);
RotateLeft(ARC);
//temp = Convert.ToString(ARC[1], 2).PadLeft(8,'0');
//MessageBox.Show(temp.ToString());
//MessageBox.Show(ByteArrayToString(ARC));
ARC[0] |= 1 << 6;
ARC[0] |= 1 << 7;
byte SB = 0x02;
string Hex = Convert.ToString(SB, 2);
Hex = Hex.Substring(0, 2);
Hex += "000000";
byte SB_processed = Convert.ToByte(Hex,2);
SB_processed |= 1 << 0;
SB_processed |= 1 << 1;
SB_processed |= 1 << 2;
SB_processed |= 1 << 3;
SB_processed |= 1 << 4;
SB_processed |= 1 << 5;
ARC[0] &= SB_processed;
MessageBox.Show(ByteArrayToString(ARC));
Logic for Left shifting taken from this reference.
c# - left shift an entire byte array

I can't well understand your question, but here I show you how to concatenate bits.
byte bits2 = 0b_000000_10;
byte bits6 = 0b_00_101100;
ushort bits14 = 0b_00_10110010001110;
uint concatenated = ((uint)bits2 << 20) + ((uint)bits6 << 14) + (uint)bits14;
uint test = 0b_00_10_101100_10110010001110;
if (concatenated == test‬)
{
Console.WriteLine("Ok");
}

C# Convert Int To Two Hex Bytes?

I have an integer that I need to convert to a four digit hex value.
For example, lets say the int value is 16. What I am looking for is a way to go from 16 to
0x00 0x10.
Any suggestions would be greatly appreciated!!!

Try this:
var input = 16;
var bytes = new byte[2];
bytes[0] = (byte)(input >> 8); // 0x00
bytes[1] = (byte)input; // 0x10
var result = (bytes[0] << 8)
| bytes[1];
// result == 16

Here's one with regular expressions, just for fun:
Regex.Replace(number.ToString("X4"), "..", "0x$0 ").TrimEnd();

Alternately, a little more general solution is to do it by byte array (then you can use this for strings or other data types)
public static string ByteArrayToString(byte[] ba)
{
string hex = BitConverter.ToString(ba);
return hex.Replace("-","");
}
int i = 39;
string str = "ssifm";
long l = 93823;
string hexi = ByteArrayToString(BitConverter.GetBytes(i));
string hexstr = ByteArrayToString(Encoding.Ascii.GetBytes(str));
string hexl = ByteArrayToString(BitConverter.GetBytes(l));
This returns them in a 'FF' format, you can add the '0x' yourself by adding this after the ToString() instead:
return "0x"+hex.Replace("-", " 0x");

Shift it! Mask it! Mask it! string.Format it!
int n = 16;
string.Format("0x{0:x2} 0x{1:x2}", (n & 0xff00) >> 8, n & 0xff); // 0x00 0x10
Here's a demo.
The x2 format specifier means a 2-digit hexadecimal value.
Okay, apparently you just want two bytes. Hexadecimal is not relevant here.
byte lowByte = (byte)(n & 0xff);
byte highByte = (byte)(n >> 8 & 0xff);

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);.

Perform signed arithmetic on numbers defined as bit ranges of unsigned bytes

I have two bytes. I need to turn them into two integers where the first 12 bits make one int and the last 4 make the other. I figure i can && the 2nd byte with 0x0f to get the 4 bits, but I'm not sure how to make that into a byte with the correct sign.
update:
just to clarify I have 2 bytes
byte1 = 0xab
byte2 = 0xcd
and I need to do something like this with it
var value = 0xabc * 10 ^ 0xd;
sorry for the confusion.
thanks for all of the help.

int a = 10;
int a1 = a&0x000F;
int a2 = a&0xFFF0;
try to use this code

For kicks:
public static partial class Levitate
{
public static Tuple<int, int> UnPack(this int value)
{
uint sign = (uint)value & 0x80000000;
int small = ((int)sign >> 28) | (value & 0x0F);
int big = value & 0xFFF0;
return new Tuple<int, int>(small, big);
}
}
int a = 10;
a.UnPack();

Ok, let's try this again knowing what we're shooting for. I tried the following out in VS2008 and it seems to work fine, that is, both outOne and outTwo = -1 at the end. Is that what you're looking for?
byte b1 = 0xff;
byte b2 = 0xff;
ushort total = (ushort)((b1 << 8) + b2);
short outOne = (short)((short)(total & 0xFFF0) >> 4);
sbyte outTwo = (sbyte)((sbyte)((total & 0xF) << 4) >> 4);

Assuming you have the following to bytes:
byte a = 0xab;
byte b = 0xcd;
and consider 0xab the first 8 bits and 0xcd the second 8 bits, or 0xabc the first 12 bits and 0xd the last four bits. Then you can get the these bits as follows;
int x = (a << 4) | (b >> 4); // x == 0x0abc
int y = b & 0x0f; // y == 0x000d

edited to take into account clarification of "signing" rules:
public void unpack( byte[] octets , out int hiNibbles , out int loNibble )
{
if ( octets == null ) throw new ArgumentNullException("octets");
if ( octets.Length != 2 ) throw new ArgumentException("octets") ;
int value = (int) BitConverter.ToInt16( octets , 0 ) ;
// since the value is signed, right shifts sign-extend
hiNibbles = value >> 4 ;
loNibble = ( value << 28 ) >> 28 ;
return ;
}