Amend 10 bits of 2 bytes - c#

In C#, how would I go about setting 2 bytes where the first 10 bits represent one decimal value and the next 6 represent a different decimal value?
So if the first value was '8' (first 10 bits) and the second '2' (remaining 6 bits), I need to end up with '0000001000 000010' inside a byte array.
Thanks!
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UInt16 val1 = 8;
UInt16 val2 = 2;
UInt16 combined = (UInt16)((val1 << 6) | val2);
If you need it in a byte array, you can pass the result to the BitConverter.GetBytes method.
byte[] array = BitConverter.GetBytes(combined);

int val1 = 8;
int val2 = 2;
// First byte contains all but the 2 least significant bits from the first value.
byte byte1 = (byte)(val1 >> 2);
// Second byte contains the 2 least significant bits from the first value,
// shifted 6 bits left to become the 2 most significant bits of the byte,
// followed by the (at most 6) bits of the second value.
byte byte2 = (byte)((val1 & 4) << 6 | val2);
byte[] bytes = new byte[] { byte1, byte2 };
// Just for verification.
string s =
Convert.ToString(byte1, 2).PadLeft(8, '0') + " " +
Convert.ToString(byte2, 2).PadLeft(8, '0');

Not accounting for any kind of overflow:
private static byte[] amend(int a, int b)
{
// Combine the datum into a 16 bits integer
var c = (ushort) ((a << 6) | (b));
// Fragment the Int to bytes
var ret = new byte[2];
ret[0] = (byte) (c >> 8);
ret[1] = (byte) (c);
return ret;
}

ushort value = (8 << 6) | 2;
byte[] bytes = BitConverter.GetBytes(value);

Related

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);
}

C# Convert int to short and then to bytes and back to int

I'm trying to convert int to short and then to byte[] but I'm getting wrong values, i pass in 1 and get 256 what am i doing wrong?
this is the code:
//passing 1
int i = 1;
byte[] shortBytes = ShortAsByte((short)i);
//ii is 256
short ii = Connection.BytesToShort (shortBytes [0], shortBytes [1]);
public static byte[] ShortAsByte(short shortValue){
byte[] intBytes = BitConverter.GetBytes(shortValue);
if (BitConverter.IsLittleEndian) Array.Reverse(intBytes);
return intBytes;
}
public static short BytesToShort(byte byte1, byte byte2)
{
return (short)((byte2 << 8) + byte1);
}
The method ShortAsByte has the most significant bit at index 0 and the least significant at index 1, so the BytesToShort method is shifting a 1 instead of a 0. This means BytesToShort returns 256 (1 << 8 + 0 = 256) instead of 1 (0 << 8 + 1 = 1) .
Swap the byte variables in the return statement to get the correct result.
public static short BytesToShort(byte byte1, byte byte2)
{
return (short)((byte1 << 8) + byte2);
}
Also, props to you for taking endian-ness into consideration!

Combine 2 numbers in a byte

I have two numbers (going from 0-9) and I want to combine them into 1 byte.
Number 1 would take bit 0-3 and Number 2 has bit 4-7.
Example : I have number 3 and 4.
3 = 0011 and 4 is 0100.
Result should be 0011 0100.
How can I make a byte with these binary values?
This is what I currently have :
public Byte CombinePinDigit(int DigitA, int DigitB)
{
BitArray Digit1 = new BitArray(Convert.ToByte(DigitA));
BitArray Digit2 = new BitArray(Convert.ToByte(DigitB));
BitArray Combined = new BitArray(8);
Combined[0] = Digit1[0];
Combined[1] = Digit1[1];
Combined[2] = Digit1[2];
Combined[3] = Digit1[3];
Combined[4] = Digit2[0];
Combined[5] = Digit2[1];
Combined[6] = Digit2[2];
Combined[7] = Digit2[3];
}
With this code I have ArgumentOutOfBoundsExceptions
Forget all that bitarray stuff.
Just do this:
byte result = (byte)(number1 | (number2 << 4));
And to get them back:
int number1 = result & 0xF;
int number2 = (result >> 4) & 0xF;
This works by using the << and >> bit-shift operators.
When creating the byte, we are shifting number2 left by 4 bits (which fills the lowest 4 bits of the results with 0) and then we use | to or those bits with the unshifted bits of number1.
When restoring the original numbers, we reverse the process. We shift the byte right by 4 bits which puts the original number2 back into its original position and then we use & 0xF to mask off any other bits.
This bits for number1 will already be in the right position (since we never shifted them) so we just need to mask off the other bits, again with & 0xF.
You should verify that the numbers are in the range 0..9 before doing that, or (if you don't care if they're out of range) you can constrain them to 0..15 by anding with 0xF:
byte result = (byte)((number1 & 0xF) | ((number2 & 0xF) << 4));
this should basically work:
byte Pack(int a, int b)
{
return (byte)(a << 4 | b & 0xF);
}
void Unpack(byte val, out int a, out int b)
{
a = val >> 4;
b = val & 0xF;
}

Read a byte starting from a specific bit in a file - C#

I need to read a byte from a .bin but starting from the specific bit, for example:
If I have this two bytes:
01010111 10101100
The program should be able to read from any bit, let's say starting in bit number 3 (or index 2):
01[010111 10]101100
And the result should be 01011110
I can read a byte starting from any bit except when the starting bit is the one at the end of a byte: 0101011[1 ...] //returns something different..
my code is:
byte readByte(int indexInBits, byte[] bytes)
{
int actualByte = (indexInBits+1)/8;
int indexInByte = (indexInBits)%8;
int b1 = bytes[actualByte] << indexInByte;
int b2 = bytes[actualByte+1] >> 8 - indexInByte;
return (byte)(b1 + b2);
}
What is wrong with it?
Thanks
byte ReadByte(int index, byte[] bytes)
{
int bytePos = index / 8;
int bitPos = index % 8;
int byte1 = bytes[bytePos] << bitPos;
int byte2 = bytes[bytePos + 1] >> 8 - bitPos;
return (byte)(byte1 + byte2);
}
I couldn't verify this right now, but this should work as expected.

C# Language: Changing the First Four Bits in a Byte

In order to utilize a byte to its fullest potential, I'm attempting to store two unique values into a byte: one in the first four bits and another in the second four bits. However, I've found that, while this practice allows for optimized memory allocation, it makes changing the individual values stored in the byte difficult.
In my code, I want to change the first set of four bits in a byte while maintaining the value of the second four bits in the same byte. While bitwise operations allow me to easily retrieve and manipulate the first four bit values, I'm finding it difficult to concatenate this new value with the second set of four bits in a byte. The question is, how can I erase the first four bits from a byte (or, more accurately, set them all the zero) and add the new set of 4 bits to replace the four bits that were just erased, thus preserving the last 4 bits in a byte while changing the first four?
Here's an example:
// Changes the first four bits in a byte to the parameter value
public void changeFirstFourBits(byte newFirstFour)
{
// If 'newFirstFour' is 0101 in binary, make 'value' 01011111 in binary, changing
// the first four bits but leaving the second four alone.
}
private byte value = 255; // binary: 11111111
Use bitwise AND (&) to clear out the old bits, shift the new bits to the correct position and bitwise OR (|) them together:
value = (value & 0xF) | (newFirstFour << 4);
Here's what happens:
value : abcdefgh
newFirstFour : 0000xyzw
0xF : 00001111
value & 0xF : 0000efgh
newFirstFour << 4 : xyzw0000
(value & 0xF) | (newFirstFour << 4) : xyzwefgh
When I have to do bit-twiddling like this, I make a readonly struct to do it for me. A four-bit integer is called nybble, of course:
struct TwoNybbles
{
private readonly byte b;
public byte High { get { return (byte)(b >> 4); } }
public byte Low { get { return (byte)(b & 0x0F); } {
public TwoNybbles(byte high, byte low)
{
this.b = (byte)((high << 4) | (low & 0x0F));
}
And then add implicit conversions between TwoNybbles and byte. Now you can just treat any byte as having a High and Low byte without putting all that ugly bit twiddling in your mainline code.
You first mask out you the high four bytes using value & 0xF. Then you shift the new bits to the high four bits using newFirstFour << 4 and finally you combine them together using binary or.
public void changeHighFourBits(byte newHighFour)
{
value=(byte)( (value & 0x0F) | (newFirstFour << 4));
}
public void changeLowFourBits(byte newLowFour)
{
value=(byte)( (value & 0xF0) | newLowFour);
}
I'm not really sure what your method there is supposed to do, but here are some methods for you:
void setHigh(ref byte b, byte val) {
b = (b & 0xf) | (val << 4);
}
byte high(byte b) {
return (b & 0xf0) >> 4;
}
void setLow(ref byte b, byte val) {
b = (b & 0xf0) | val;
}
byte low(byte b) {
return b & 0xf;
}
Should be self-explanatory.
public int SplatBit(int Reg, int Val, int ValLen, int Pos)
{
int mask = ((1 << ValLen) - 1) << Pos;
int newv = Val << Pos;
int res = (Reg & ~mask) | newv;
return res;
}
Example:
Reg = 135
Val = 9 (ValLen = 4, because 9 = 1001)
Pos = 2
135 = 10000111
9 = 1001
9 << Pos = 100100
Result = 10100111
A quick look would indicate that a bitwise and can be achieved using the & operator. So to remove the first four bytes you should be able to do:
byte value1=255; //11111111
byte value2=15; //00001111
return value1&value2;
Assuming newVal contains the value you want to store in origVal.
Do this for the 4 least significant bits:
byte origVal = ???;
byte newVal = ???
orig = (origVal & 0xF0) + newVal;
and this for the 4 most significant bits:
byte origVal = ???;
byte newVal = ???
orig = (origVal & 0xF) + (newVal << 4);
I know you asked specifically about clearing out the first four bits, which has been answered several times, but I wanted to point out that if you have two values <= decimal 15, you can combine them into 8 bits simply with this:
public int setBits(int upperFour, int lowerFour)
{
return upperFour << 4 | lowerFour;
}
The result will be xxxxyyyy where
xxxx = upperFour
yyyy = lowerFour
And that is what you seem to be trying to do.
Here's some code, but I think the earlier answers will do it for you. This is just to show some sort of test code to copy and past into a simple console project (the WriteBits method by be of help):
static void Main(string[] args)
{
int b1 = 255;
WriteBits(b1);
int b2 = b1 >> 4;
WriteBits(b2);
int b3 = b1 & ~0xF ;
WriteBits(b3);
// Store 5 in first nibble
int b4 = 5 << 4;
WriteBits(b4);
// Store 8 in second nibble
int b5 = 8;
WriteBits(b5);
// Store 5 and 8 in first and second nibbles
int b6 = 0;
b6 |= (5 << 4) + 8;
WriteBits(b6);
// Store 2 and 4
int b7 = 0;
b7 = StoreFirstNibble(2, b7);
b7 = StoreSecondNibble(4, b7);
WriteBits(b7);
// Read First Nibble
int first = ReadFirstNibble(b7);
WriteBits(first);
// Read Second Nibble
int second = ReadSecondNibble(b7);
WriteBits(second);
}
static int ReadFirstNibble(int storage)
{
return storage >> 4;
}
static int ReadSecondNibble(int storage)
{
return storage &= 0xF;
}
static int StoreFirstNibble(int val, int storage)
{
return storage |= (val << 4);
}
static int StoreSecondNibble(int val, int storage)
{
return storage |= val;
}
static void WriteBits(int b)
{
Console.WriteLine(BitConverter.ToString(BitConverter.GetBytes(b),0));
}
}

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