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c# FTDI 24bit two's complement implement

I am a beginner at C# and would like some advice on how to solve the following problem:
My code read 3 byte from ADC true FTDI ic and calculating value. Problem is that sometimes I get values under zero (less than 0000 0000 0000 0000 0000 0000) and my calculating value jump to a big number. I need to implement two's complement but I and I don't know how.
byte[] readData1 = new byte[3];
ftStatus = myFtdiDevice.Read(readData1, numBytesAvailable, ref numBytesRead); //read data from FTDI
int vrednost1 = (readData1[2] << 16) | (readData1[1] << 8) | (readData1[0]); //convert LSB MSB
int v = ((((4.096 / 16777216) * vrednost1) / 4) * 250); //calculate

Convert the data left-justified, so that your sign-bit lands in the spot your PC processor treats as a sign bit. Then right-shift back, which will automatically perform sign-extension.
int left_justified = (readData[2] << 24) | (readData[1] << 16) | (readData[0] << 8);
int result = left_justified >> 8;
Equivalently, one can mark the byte containing a sign bit as signed, so the processor will perform sign extension:
int result = (unchecked((sbyte)readData[2]) << 16) | (readData[1] << 8) | readData[0];
The second approach with a cast only works if the sign bit is already aligned left within any one of the bytes. The left-justification approach can work for arbitrary sizes, such as 18-bit two's complement readings. In this situation the cast to sbyte wouldn't do the job.
int left_justified18 = (readData[2] << 30) | (readData[1] << 22) | (readData[0] << 14);
int result18 = left_justified >> 14;

Well, the only obscure moment here is endianness (big or little). Assuming that byte[0] stands for the least significant byte you can put
private static int FromInt24(byte[] data) {
int result = (data[2] << 16) | (data[1] << 8) | data[0];
return data[2] < 128
? result // positive number
: -((~result & 0xFFFFFF) + 1); // negative number, its abs value is 2-complement
}
Demo:
byte[][] tests = new byte[][] {
new byte[] { 255, 255, 255},
new byte[] { 44, 1, 0},
};
string report = string.Join(Environment.NewLine, tests
.Select(test => $"[{string.Join(", ", test.Select(x => $"0x{x:X2}"))}] == {FromInt24(test)}"));
Console.Write(report);
Outcome:
[0xFF, 0xFF, 0xFF] == -1
[0x2C, 0x01, 0x00] == 300
If you have big endianness (e.g. 300 == {0, 1, 44}) you have to swap bytes:
private static int FromInt24(byte[] data) {
int result = (data[0] << 16) | (data[1] << 8) | data[2];
return data[0] < 128
? result
: -((~result & 0xFFFFFF) + 1);
}

Why BinaryReader.ReadInt64() read data as upper/lower separately?

I have questions for BinaryReader by Microsoft.
From https://referencesource.microsoft.com/#mscorlib/system/io/binaryreader.cs ,
public virtual long ReadInt64() {
FillBuffer(8);
uint lo = (uint)(m_buffer[0] | m_buffer[1] << 8 |
m_buffer[2] << 16 | m_buffer[3] << 24);
uint hi = (uint)(m_buffer[4] | m_buffer[5] << 8 |
m_buffer[6] << 16 | m_buffer[7] << 24);
return (long) ((ulong)hi) << 32 | lo;
}
It parse data as two parts (upper 32bit, lower 32bit) for Int64 data. But I think that it is rebundant. So the method like below also should work.
public virtual long ReadInt64Other() {
FillBuffer(8);
return (long)m_buffer[0] |
(long)m_buffer[1] << 8 |
(long)m_buffer[2] << 16 |
(long)m_buffer[3] << 24 |
(long)m_buffer[4] << 32 |
(long)m_buffer[5] << 40 |
(long)m_buffer[6] << 48 |
(long)m_buffer[7] << 56;
}
Is there any reason for Microsoft use upper/lower style parsing?

Get integer from last 12 bits of 2 bytes in C#

I suspect this is an easy one.
I need to get a number from the first 4 bits and another number from the last 12 bits
of 2 bytes.
So here is what I have but it doesn't seem to be right:
byte[] data = new byte[2];
//assume byte array contains data
var _4bit = data[0] >> 4;
var _12bit = data[0] >> 8 | data[1] & 0xff;

data[0]>>8 is 0. Remember that your data is defined as byte[] so it has 8bits per single item, so you are effectively cutting ALL bits off the data[0].
You want rather to take the lowest 4 bits from that byte by bitwise AND (00001111 = 0F) and then shift it leftwards as needed.
So try this:
var _4bit = data[0] >> 4;
var _12bit = ((data[0] & 0x0F) << 8) | (data[1] & 0xff);
It's also worth noting that the last & 0xFF is not needed, as the data[1] is already a byte.
On bits, step by step:
byte[2] data = { aaaabbbb, cccccccc }
var _4bit = data[0] >> 4;
= aaaabbbb >> 4
= 0000aaaa
var _12bit = ( (data[0] & 0x0F) << 8) | ( data[1] & 0xff);
= ((aaaabbbb & 0x0F) << 8) | (cccccccc & 0xff);
= ( 0000bbbb << 8) | ( cccccccc );
= ( 0000bbbb000000000 ) | ( cccccccc );
= 0000bbbbcccccccc;
BTW. also, note that results of & and | operators are typed as int, so 32bits, I've omitted the zeroes for clarity and written it as 8bit only to make it brief!

Bitshift over Int32.max fails

Ok, so I have two methods
public static long ReadLong(this byte[] data)
{
if (data.Length < 8) throw new ArgumentOutOfRangeException("Not enough data");
long length = data[0] | data[1] << 8 | data[2] << 16 | data[3] << 24
| data[4] << 32 | data[5] << 40 | data[6] << 48 | data[7] << 56;
return length;
}
public static void WriteLong(this byte[] data, long i)
{
if (data.Length < 8) throw new ArgumentOutOfRangeException("Not enough data");
data[0] = (byte)((i >> (8*0)) & 0xFF);
data[1] = (byte)((i >> (8*1)) & 0xFF);
data[2] = (byte)((i >> (8*2)) & 0xFF);
data[3] = (byte)((i >> (8*3)) & 0xFF);
data[4] = (byte)((i >> (8*4)) & 0xFF);
data[5] = (byte)((i >> (8*5)) & 0xFF);
data[6] = (byte)((i >> (8*6)) & 0xFF);
data[7] = (byte)((i >> (8*7)) & 0xFF);
}
So WriteLong works correctly(Verified against BitConverter.GetBytes()). The problem is ReadLong. I have a fairly good understanding of this stuff, but I'm guessing what's happening is the or operations are happening as 32 bit ints so at Int32.MaxValue it rolls over. I'm not sure how to avoid that. My first instinct was to make an int from the lower half and an int from the upper half and combine them, but I'm not quite knowledgeable to know even where to start with that, so this is what I tried....
public static long ReadLong(byte[] data)
{
if (data.Length < 8) throw new ArgumentOutOfRangeException("Not enough data");
long l1 = data[0] | data[1] << 8 | data[2] << 16 | data[3] << 24;
long l2 = data[4] | data[5] << 8 | data[6] << 16 | data[7] << 24;
return l1 | l2 << 32;
}
This didn't work though, at least not for larger numbers, it seems to work for everything below zero.
Here's how I run it
void Main()
{
var larr = new long[5]{
long.MinValue,
0,
long.MaxValue,
1,
-2000000000
};
foreach(var l in larr)
{
var arr = new byte[8];
WriteLong(ref arr,l);
Console.WriteLine(ByteString(arr));
var end = ReadLong(arr);
var end2 = BitConverter.ToInt64(arr,0);
Console.WriteLine(l + " == " + end + " == " + end2);
}
}
and here's what I get(using the modified ReadLong method)
0:0:0:0:0:0:0:128
-9223372036854775808 == -9223372036854775808 == -9223372036854775808
0:0:0:0:0:0:0:0
0 == 0 == 0
255:255:255:255:255:255:255:127
9223372036854775807 == -1 == 9223372036854775807
1:0:0:0:0:0:0:0
1 == 1 == 1
0:108:202:136:255:255:255:255
-2000000000 == -2000000000 == -2000000000

The problem is not the or, it is the bitshift. This has to be done as longs. Currently, the data[i] are implicitely converted to int. Just change that to long and that's it. I.e.
public static long ReadLong(byte[] data)
{
if (data.Length < 8) throw new ArgumentOutOfRangeException("Not enough data");
long length = (long)data[0] | (long)data[1] << 8 | (long)data[2] << 16 | (long)data[3] << 24
| (long)data[4] << 32 | (long)data[5] << 40 | (long)data[6] << 48 | (long)data[7] << 56;
return length;
}

You are doing int arithmetic and then assigning to long, try:
long length = data[0] | data[1] << 8L | data[2] << 16L | data[3] << 24L
| data[4] << 32L | data[5] << 40L | data[6] << 48L | data[7] << 56L;
This should define your constants as longs forcing it to use long arithmetic.
EDIT: Turns out this may not work according to comments below as while bitshift takes many operators on the left it only takes int on the right. Georg's should be the accepted answer.

C# Bitwise Operators

Could someone please explain what the following code does.
private int ReadInt32(byte[] _il, ref int position)
{
return (((il[position++] | (il[position++] << 8)) | (il[position++] << 0x10)) | (il[position++] << 0x18));
}
I'm not sure I understand how the bitwise operators in this method work, could some please break it down for me?

The integer is given as a byte array.
Then each byte is shifted left 0/8/16/24 places and these values are summed to get the integer value.
This is an Int32 in hexadecimal format:
0x10203040
It is represented as following byte array (little endian architecture, so bytes are in reverse order):
[0x40, 0x30, 0x20, 0x10]
In order to build the integer back from the array, each element is shifted i.e. following logic is performed:
a = 0x40 = 0x00000040
b = 0x30 << 8 = 0x00003000
c = 0x20 << 16 = 0x00200000
d = 0x10 << 24 = 0x10000000
then these values are OR'ed together:
int result = a | b | c | d;
this gives:
0x00000040 |
0x00003000 |
0x00200000 |
0x10000000 |
------------------
0x10203040

Think of it like this:
var i1 = il[position];
var i2 = il[position + 1] << 8; (<< 8 is equivalent to * 256)
var i3 = il[position + 2] << 16;
var i4 = il[position + 3] << 24;
position = position + 4;
return i1 | i2 | i3 | i4;