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byte array from specific index as struct in c# without making a copy

Currently I code client-server junk and deal a lot with C++ structs passed over network.
I know about ways provided here Reading a C/C++ data structure in C# from a byte array, but they all about making a copy.
I want to have something like that:
struct/*or class*/ SomeStruct
{
public uint F1;
public uint F2;
public uint F3;
}
Later in my code I want to have something like that:
byte[] Data; //16 bytes that I got from network
SomeStruct PartOfDataAsSomeStruct { get { return /*make SomeStruct instance based on this.Data starting from index 4, without copying it. So when I do PartOfDataAsSomeStruct.F1 = 132465; it also changes bytes 4, 5, 6 and 7 in this.Data.*/; } }
If this is possible, please, tell how?

Like so?
byte[] data = new byte[16];
// 00-00-00-00-00-00-00-00-00-00-00-00-00-00-00-00
Console.WriteLine(BitConverter.ToString(data));
ref SomeStruct typed = ref Unsafe.As<byte, SomeStruct>(ref data[4]);
typed.F1 = 42;
typed.F2 = 3;
typed.F3 = 9;
// 00-00-00-00-2A-00-00-00-03-00-00-00-09-00-00-00
Console.WriteLine(BitConverter.ToString(data));
This coerces the data from the middle of the byte-array using a ref-local that is an "interior managed pointer" to the data. Zero copies.
If you need multiple items (like how a vector would work), you can do the same thing with spans and MemoryMarshal.Cast
Note that it uses CPU-endian rules for the elements - little endian in my case.
For spans:
byte[] data = new byte[256];
// create a span of some of it
var span = new Span<byte>(data, 4, 128);
// now coerce the span
var typed = MemoryMarshal.Cast<byte, SomeStruct>(span);
Console.WriteLine(typed.Length); // 10 of them fit
typed[3].F1 = 3; // etc

Thank you for the correction, Marc Gravell. And thank you for the example.
Here is a way using Class and Bitwise Operators, without pointers, to do the samething:
class SomeClass
{
public byte[] Data;
public SomeClass()
{
Data = new byte[16];
}
public uint F1
{
get
{
uint ret = (uint)(Data[4] << 24 | Data[5] << 16 | Data[6] << 8 | Data[7]);
return ret;
}
set
{
Data[4] = (byte)(value >> 24);
Data[5] = (byte)(value >> 16);
Data[6] = (byte)(value >> 8);
Data[7] = (byte)value;
}
}
}
Testing:
SomeClass sc = new SomeClass();
sc.F1 = 0b_00000001_00000010_00000011_00000100;
Console.WriteLine(sc.Data[3].ToString() + " " + sc.Data[4].ToString() + " " + sc.Data[5].ToString() + " " + sc.Data[6].ToString());
Console.WriteLine(sc.F1.ToString());
//Output:
//1 2 3 4
//16909060

Refactor byte array into ENUM values

I have a function that receives some data I have to respond with a HEX value.
public byte[] GetData(string value)
{
byte[] returnVal = null;
switch(value){
case "demo1":
byte byte1 = (byte)((32769 & 0x000000FF));
byte byte2 = (byte)((32769 & 0x0000FF00) >> 8);
returnVal = new byte[] { byte1, byte2 };
break;
.....
}
return returnVal;
}
In this example I have to response with 0x8001
I create the following code to build manually a 2 byte array with the right response.
byte byte1 = (byte)((32769 & 0x000000FF));
byte byte2 = (byte)((32769 & 0x0000FF00) >> 8);
var resCmd = new byte[] { byte1, byte2 };
The response can be different depending on the value I received so I want to have a ENUM with the responses and then convert that to byte array.
For example:
public enum Commands
{
CMD1 = 0x8001,
CMD2 = 0x8002,
CMD3 = 0x8003
};
How can I convert an Enum for example CMD1 to the 2 byte array that I need?
Thanks

Use BitConverter after casting it to a 16 bit unsigned int:
var bytes = BitConverter.GetBytes((UInt16)Commands.CMD1);

int to byte[] consistency over network

I have a struct that gets used all over the place and that I store as byteArray on the hd and also send to other platforms.
I used to do this by getting a string version of the struct and using getBytes(utf-8) and getString(utf-8) during serialization. With that I guess I avoided the little and big endian problems?
However that was quite a bit of overhead and I am now using this:
public static explicit operator byte[] (Int3 self)
{
byte[] int3ByteArr = new byte[12];//4*3
int x = self.x;
int3ByteArr[0] = (byte)x;
int3ByteArr[1] = (byte)(x >> 8);
int3ByteArr[2] = (byte)(x >> 0x10);
int3ByteArr[3] = (byte)(x >> 0x18);
int y = self.y;
int3ByteArr[4] = (byte)y;
int3ByteArr[5] = (byte)(y >> 8);
int3ByteArr[6] = (byte)(y >> 0x10);
int3ByteArr[7] = (byte)(y >> 0x18);
int z = self.z;
int3ByteArr[8] = (byte)z;
int3ByteArr[9] = (byte)(z >> 8);
int3ByteArr[10] = (byte)(z >> 0x10);
int3ByteArr[11] = (byte)(z >> 0x18);
return int3ByteArr;
}
public static explicit operator Int3(byte[] self)
{
int x = self[0] + (self[1] << 8) + (self[2] << 0x10) + (self[3] << 0x18);
int y = self[4] + (self[5] << 8) + (self[6] << 0x10) + (self[7] << 0x18);
int z = self[8] + (self[9] << 8) + (self[10] << 0x10) + (self[11] << 0x18);
return new Int3(x, y, z);
}
It works quite well for me, but I am not quite sure how little/big endian works,. do I still have to take care of something here to be safe when some other machine receives an int I sent as a bytearray?

Your current approach will not work for the case when your application running on system which use Big-Endian. In this situation you don't need reordering at all.
You don't need to reverse byte arrays by your self
And you don't need check for endianess of the system by your self
Static method IPAddress.HostToNetworkOrder will convert integer to the integer with big-endian order.
Static method IPAddress.NetworkToHostOrder will convert integer to the integer with order your system using
Those methods will check for Endianness of the system and will do/or not reordering of integers.
For getting bytes from integer and back use BitConverter
public struct ThreeIntegers
{
public int One;
public int Two;
public int Three;
}
public static byte[] ToBytes(this ThreeIntegers value )
{
byte[] bytes = new byte[12];
byte[] bytesOne = IntegerToBytes(value.One);
Buffer.BlockCopy(bytesOne, 0, bytes, 0, 4);
byte[] bytesTwo = IntegerToBytes(value.Two);
Buffer.BlockCopy(bytesTwo , 0, bytes, 4, 4);
byte[] bytesThree = IntegerToBytes(value.Three);
Buffer.BlockCopy(bytesThree , 0, bytes, 8, 4);
return bytes;
}
public static byte[] IntegerToBytes(int value)
{
int reordered = IPAddress.HostToNetworkOrder(value);
return BitConverter.GetBytes(reordered);
}
And converting from bytes to struct
public static ThreeIntegers GetThreeIntegers(byte[] bytes)
{
int rawValueOne = BitConverter.ToInt32(bytes, 0);
int valueOne = IPAddress.NetworkToHostOrder(rawValueOne);
int rawValueTwo = BitConverter.ToInt32(bytes, 4);
int valueTwo = IPAddress.NetworkToHostOrder(rawValueTwo);
int rawValueThree = BitConverter.ToInt32(bytes, 8);
int valueThree = IPAddress.NetworkToHostOrder(rawValueThree);
return new ThreeIntegers(valueOne, valueTwo, valueThree);
}
If you will use BinaryReader and BinaryWriter for saving and sending to another platforms then BitConverter and byte array manipulating can be dropped off.
// BinaryWriter.Write have overload for Int32
public static void SaveThreeIntegers(ThreeIntegers value)
{
using(var stream = CreateYourStream())
using (var writer = new BinaryWriter(stream))
{
int reordredOne = IPAddress.HostToNetworkOrder(value.One);
writer.Write(reorderedOne);
int reordredTwo = IPAddress.HostToNetworkOrder(value.Two);
writer.Write(reordredTwo);
int reordredThree = IPAddress.HostToNetworkOrder(value.Three);
writer.Write(reordredThree);
}
}
For reading value
public static ThreeIntegers LoadThreeIntegers()
{
using(var stream = CreateYourStream())
using (var writer = new BinaryReader(stream))
{
int rawValueOne = reader.ReadInt32();
int valueOne = IPAddress.NetworkToHostOrder(rawValueOne);
int rawValueTwo = reader.ReadInt32();
int valueTwo = IPAddress.NetworkToHostOrder(rawValueTwo);
int rawValueThree = reader.ReadInt32();
int valueThree = IPAddress.NetworkToHostOrder(rawValueThree);
}
}
Of course you can refactor methods above and get more cleaner solution.
Or add as extension methods for BinaryWriter and BinaryReader.

Yes you do. With changes endianness your serialization which preserves bit ordering will run into trouble.
Take the int value 385
In a bigendian system it would be stored as
000000000000000110000001
Interpreting it as littleendian would read it as
100000011000000000000000
And reverse translate to 8486912
If you use the BitConverter class there will be a book property desiring the endianness of the system. The bitconverter can also produce the bit arrays for you.
You will have to decide to use either endianness and reverse the byte arrays according to the serializing or deserializing systems endianness.
The description on MSDN is actually quite detailed. Here they use Array.Reverse for simplicity. I am not certain that your casting to/from byte in order to do the bit manipulation is in fact the fastest way of converting, but that is easily benchmarked.

How do i create byte array that contains 64 bits array and how do i convert those bits into hex value?

I want to create byte array that contains 64 bits, How can i get particular bits values say 17th bit, and also how can i get hex value of that index of byte? I did like this, Is this correct?
byte[] _byte = new byte[8];
var bit17=((((_byte[2]>>1)& 0x01);
string hex=BitConverter.ToString(_byte,2,4).Replace("-", string.Empty)

You could use a BitArray:
var bits = new BitArray(64);
bool bit17 = bits[17];
I'm not sure what you mean by the "hex value of that bit" - it will be 0 or 1, because it's a bit.
If you have the index of a bit in a byte (between 0 and 7 inclusive) then you can convert that to a hex string as follows:
int bitNumber = 7; // For example.
byte value = (byte)(1 << bitNumber);
string hex = value.ToString("x");
Console.WriteLine(hex);

You can just use ToString() method.
byte[] arr= new byte[8];
int index = 0;
string hexValue = arr[index].ToString("X");

Converting 2 bytes to Short in C#

I'm trying to convert two bytes into an unsigned short so I can retrieve the actual server port value. I'm basing it off from this protocol specification under Reply Format. I tried using BitConverter.ToUint16() for this, but the problem is, it doesn't seem to throw the expected value. See below for a sample implementation:
int bytesRead = 0;
while (bytesRead < ms.Length)
{
int first = ms.ReadByte() & 0xFF;
int second = ms.ReadByte() & 0xFF;
int third = ms.ReadByte() & 0xFF;
int fourth = ms.ReadByte() & 0xFF;
int port1 = ms.ReadByte();
int port2 = ms.ReadByte();
int actualPort = BitConverter.ToUInt16(new byte[2] {(byte)port1 , (byte)port2 }, 0);
string ip = String.Format("{0}.{1}.{2}.{3}:{4}-{5} = {6}", first, second, third, fourth, port1, port2, actualPort);
Debug.WriteLine(ip);
bytesRead += 6;
}
Given one sample data, let's say for the two byte values, I have 105 & 135, the expected port value after conversion should be 27015, but instead I get a value of 34665 using BitConverter.
Am I doing it the wrong way?

If you reverse the values in the BitConverter call, you should get the expected result:
int actualPort = BitConverter.ToUInt16(new byte[2] {(byte)port2 , (byte)port1 }, 0);
On a little-endian architecture, the low order byte needs to be second in the array. And as lasseespeholt points out in the comments, you would need to reverse the order on a big-endian architecture. That could be checked with the BitConverter.IsLittleEndian property. Or it might be a better solution overall to use IPAddress.HostToNetworkOrder (convert the value first and then call that method to put the bytes in the correct order regardless of the endianness).

BitConverter is doing the right thing, you just have low-byte and high-byte mixed up - you can verify using a bitshift manually:
byte port1 = 105;
byte port2 = 135;
ushort value = BitConverter.ToUInt16(new byte[2] { (byte)port1, (byte)port2 }, 0);
ushort value2 = (ushort)(port1 + (port2 << 8)); //same output

To work on both little and big endian architectures, you must do something like:
if (BitConverter.IsLittleEndian)
actualPort = BitConverter.ToUInt16(new byte[2] {(byte)port2 , (byte)port1 }, 0);
else
actualPort = BitConverter.ToUInt16(new byte[2] {(byte)port1 , (byte)port2 }, 0);