Related
Let say we have two bitmaps that are represented by unsigned long(64-bit) arrays. And I want to merge this two bitmaps using specific shift(offset).
For example merge bitmap1(bigger) into bitmap2(smaller) starting offset 3. Offset 3 mean that 3rd bit of bitmap1 corresponds to 0 bit of bitmap2.
By merge I mean logical Or operation. What is the cleanest way to do this?
Currently I have done this with simple uneffective for loop
const ulong BitsPerUlong = 64;
MergeAt(ulong startIndex, Bitmap bitmap2)
{
for (int i = startIndex; i < bitmap2.Capacity; i++)
{
bool newVal = bitmap2.GetAt(i) | bitmap1.GetAt(i)
bitmap2.SetAt(i, newVal)
}
}
bool GetAt(ulong index)
{
var dataOffset = BitOffsetToUlongOffset(index);
ulong mask = 0x1ul << ((int)(index % BitsPerUlong));
return (_data[dataOffset] & mask) == mask;
}
void SetAt(ulong index, bool value)
{
var dataOffset = BitOffsetToUlongOffset(index);
ulong mask = 0x1ul << ((int)(index % BitsPerUlong));
if (value)
{
_data[dataOffset] |= mask;
}
else
{
_data[dataOffset] &= ~mask;
}
}
ulong BitOffsetToUlongOffset(ulong index)
{
var dataOffset = index / BitsPerUlong;
return dataOffset;
}
(C/C++/C# accepted).
As you probably figured out yourself, if offset < BitsPerULong the first block can be merged with:
data1[0] |= data2[0] << offset;
Which leaves some bits in data2[0] unmerged, but you can get those with:
data2[0] >> (BitsPerULong - offset)
So the next merge for i > 0 becomes:
data1[i] |= (data2[i] << offset) | (data2[i-1] >> (BitsPerULong - offset));
from which you can construct a for-loop to merge all data. Of course, this still means a couple of bits from data2 will "fall off" but I think that's inherent to your problem description?
If you need a more generic solution where offset can also be greater than BitsPerULong, this needs a bit more work.
I presume you mean that you want to "merge" the smaller INTO the bigger.
Have you tried: bitmapLarger |= ( bitmapSmaller << 3 ) ?
So I need help converting this ada code int c#, it's basically a checksum algorithm.
ADA:
CHECKSUM_VALUE := ((ROTATE_LEFT_1_BIT(CHECKSUM_VALUE)) xor (CURRENT_VALUE));
This is what I could come up with:
C#:
checksum = RotateLeft(checksum, rotateCount, sizeof(ushort) * 8) ^ word;
RotateLeft Function:
public static int RotateLeft(int value, ushort rotateCount, int dataSize)
{
return (value << rotateCount) | (value >> (dataSize - rotateCount));
}
However when comparing the checksum results from the ada and C# algorithms, they do not match so I think my conversion isn't correct, anyone who has used ada before can give some input would be really helpful.
Thanks
The issue seems to be with the C# and perhaps not with your interpretation of the ADA code. If you are truly rotating a 16 bit unsigned number as your post is implying, then you will need to mask the upper 2 bytes of the resulting integer value so that they do not contribute to the answer. Casting an uint x to ushort in C# will do the equivalent of x & 0x0000FFFF
public static ushort RotateLeft(ushort value, int count)
{
int left = value << count;
int right = value >> (16 - count);
return (ushort)(left | right);
}
This answer is in C, since I don’t have a C# compiler.
You have value as an int, which is signed, so that a right shift extends the sign bit into the vacated space; so in (value << rotateCount) | (value >> (dataSize - rotateCount)), the right-hand half ((value >> (dataSize - rotateCount))) needs to have the top bits masked off. And I don’t know why you need dataSize, isn’t it sizeof(value)?
I think a better solution would be to use unsigned, so that a right shift introduces zeros into the vacated space.
#include <stdio.h>
unsigned rotateLeft(unsigned value, int by) {
const unsigned bits = sizeof(value) * 8;
return (value << by) | (value >> (bits - by));
}
int main() {
unsigned input = 0x52525252;
unsigned result = input;
printf("input: %x\n", input);
{
int j;
for (j = 0; j < 8; j++) {
result = rotateLeft(result, 1);
printf("result: %x\n", result);
}
}
return 0;
}
The output is
input: 52525252
result: a4a4a4a4
result: 49494949
result: 92929292
result: 25252525
result: 4a4a4a4a
result: 94949494
result: 29292929
result: 52525252
I am making application in C# which has a byte array containing hex values.
I am getting data as a big-endian but I want it as a little-endian and I am using Bitconverter.toInt32 method for converting that value to integer.
My problem is that before converting the value, I have to copy that 4 byte data into temporary array from source byte array and then reverse that temporary byte array.
I can't reverse source array because it also contains other data.
Because of that my application becomes slow.
In the code I have one source array of byte as waveData[] which contains a lot of data.
byte[] tempForTimestamp=new byte[4];
tempForTimestamp[0] = waveData[290];
tempForTimestamp[1] = waveData[289];
tempForTimestamp[2] = waveData[288];
tempForTimestamp[3] = waveData[287];
int number = BitConverter.ToInt32(tempForTimestamp, 0);
Is there any other method for that conversion?
Add a reference to System.Memory nuget and use BinaryPrimitives.ReverseEndianness().
using System.Buffers.Binary;
number = BinaryPrimitives.ReverseEndianness(number);
It supports both signed and unsigned integers (byte/short/int/long).
In modern-day Linq the one-liner and easiest to understand version would be:
int number = BitConverter.ToInt32(waveData.Skip(286).Take(4).Reverse().ToArray(), 0);
You could also...
byte[] tempForTimestamp = new byte[4];
Array.Copy(waveData, 287, tempForTimestamp, 0, 4);
Array.Reverse(tempForTimestamp);
int number = BitConverter.ToInt32(tempForTimestamp);
:)
If you know the data is big-endian, perhaps just do it manually:
int value = (buffer[i++] << 24) | (buffer[i++] << 16)
| (buffer[i++] << 8) | buffer[i++];
this will work reliably on any CPU, too. Note i is your current offset into the buffer.
Another approach would be to shuffle the array:
byte tmp = buffer[i+3];
buffer[i+3] = buffer[i];
buffer[i] = tmp;
tmp = buffer[i+2];
buffer[i+2] = buffer[i+1];
buffer[i+1] = tmp;
int value = BitConverter.ToInt32(buffer, i);
i += 4;
I find the first immensely more readable, and there are no branches / complex code, so it should work pretty fast too. The second could also run into problems on some platforms (where the CPU is already running big-endian).
Here you go
public static int SwapEndianness(int value)
{
var b1 = (value >> 0) & 0xff;
var b2 = (value >> 8) & 0xff;
var b3 = (value >> 16) & 0xff;
var b4 = (value >> 24) & 0xff;
return b1 << 24 | b2 << 16 | b3 << 8 | b4 << 0;
}
Declare this class:
using static System.Net.IPAddress;
namespace BigEndianExtension
{
public static class BigEndian
{
public static short ToBigEndian(this short value) => HostToNetworkOrder(value);
public static int ToBigEndian(this int value) => HostToNetworkOrder(value);
public static long ToBigEndian(this long value) => HostToNetworkOrder(value);
public static short FromBigEndian(this short value) => NetworkToHostOrder(value);
public static int FromBigEndian(this int value) => NetworkToHostOrder(value);
public static long FromBigEndian(this long value) => NetworkToHostOrder(value);
}
}
Example, create a form with a button and a multiline textbox:
using BigEndianExtension;
private void button1_Click(object sender, EventArgs e)
{
short int16 = 0x1234;
int int32 = 0x12345678;
long int64 = 0x123456789abcdef0;
string text = string.Format("LE:{0:X4}\r\nBE:{1:X4}\r\n", int16, int16.ToBigEndian());
text += string.Format("LE:{0:X8}\r\nBE:{1:X8}\r\n", int32, int32.ToBigEndian());
text += string.Format("LE:{0:X16}\r\nBE:{1:X16}\r\n", int64, int64.ToBigEndian());
textBox1.Text = text;
}
//Some code...
The most straightforward way is to use the BinaryPrimitives.ReadInt32BigEndian(ReadOnlySpan) Method introduced in .NET Standard 2.1
var number = BinaryPrimitives.ReadInt32BigEndian(waveData[297..291]);
If you won't ever again need that reversed, temporary array, you could just create it as you pass the parameter, instead of making four assignments. For example:
int i = 287;
int value = BitConverter.ToInt32({
waveData(i + 3),
waveData(i + 2),
waveData(i + 1),
waveData(i)
}, 0);
I use the following helper functions
public static Int16 ToInt16(byte[] data, int offset)
{
if (BitConverter.IsLittleEndian)
return BitConverter.ToInt16(BitConverter.IsLittleEndian ? data.Skip(offset).Take(2).Reverse().ToArray() : data, 0);
return BitConverter.ToInt16(data, offset);
}
public static Int32 ToInt32(byte[] data, int offset)
{
if (BitConverter.IsLittleEndian)
return BitConverter.ToInt32(BitConverter.IsLittleEndian ? data.Skip(offset).Take(4).Reverse().ToArray() : data, 0);
return BitConverter.ToInt32(data, offset);
}
public static Int64 ToInt64(byte[] data, int offset)
{
if (BitConverter.IsLittleEndian)
return BitConverter.ToInt64(BitConverter.IsLittleEndian ? data.Skip(offset).Take(8).Reverse().ToArray() : data, 0);
return BitConverter.ToInt64(data, offset);
}
You can also use Jon Skeet "Misc Utils" library, available at https://jonskeet.uk/csharp/miscutil/
His library has many utility functions. For Big/Little endian conversions you can check the MiscUtil/Conversion/EndianBitConverter.cs file.
var littleEndianBitConverter = new MiscUtil.Conversion.LittleEndianBitConverter();
littleEndianBitConverter.ToInt64(bytes, offset);
var bigEndianBitConverter = new MiscUtil.Conversion.BigEndianBitConverter();
bigEndianBitConverter.ToInt64(bytes, offset);
His software is from 2009 but I guess it's still relevant.
I dislike BitConverter, because (as Marc Gravell answered) it is specced to rely on system endianness, meaning you technically have to do a system endianness check every time you use BitConverter to ensure you don't have to reverse the array. And usually, with saved files, you generally know the endianness you're trying to read, and that might not be the same. You might just be handling file formats with big-endian values, too, like, for instance, PNG chunks.
Because of that, I just wrote my own methods for this, which take a byte array, the read offset and read length as arguments, as well as a boolean to specify the endianness handling, and which uses bit shifting for efficiency:
public static UInt64 ReadIntFromByteArray(Byte[] data, Int32 startIndex, Int32 bytes, Boolean littleEndian)
{
Int32 lastByte = bytes - 1;
if (data.Length < startIndex + bytes)
throw new ArgumentOutOfRangeException("startIndex", "Data array is too small to read a " + bytes + "-byte value at offset " + startIndex + ".");
UInt64 value = 0;
for (Int32 index = 0; index < bytes; index++)
{
Int32 offs = startIndex + (littleEndian ? index : lastByte - index);
value |= (((UInt64)data[offs]) << (8 * index));
}
return value;
}
This code can handle any value between 1 and 8 bytes, both little-endian and big-endian. The only small usage peculiarity is that you need to both give the amount of bytes to read, and need to specifically cast the result to the type you want.
Example from some code where I used it to read the header of some proprietary image type:
Int16 imageWidth = (Int16) ReadIntFromByteArray(fileData, hdrOffset, 2, true);
Int16 imageHeight = (Int16) ReadIntFromByteArray(fileData, hdrOffset + 2, 2, true);
This will read two consecutive 16-bit integers off an array, as signed little-endian values. You can of course just make a bunch of overload functions for all possibilities, like this:
public Int16 ReadInt16FromByteArrayLe(Byte[] data, Int32 startIndex)
{
return (Int16) ReadIntFromByteArray(data, startIndex, 2, true);
}
But personally I didn't bother with that.
And, here's the same for writing bytes:
public static void WriteIntToByteArray(Byte[] data, Int32 startIndex, Int32 bytes, Boolean littleEndian, UInt64 value)
{
Int32 lastByte = bytes - 1;
if (data.Length < startIndex + bytes)
throw new ArgumentOutOfRangeException("startIndex", "Data array is too small to write a " + bytes + "-byte value at offset " + startIndex + ".");
for (Int32 index = 0; index < bytes; index++)
{
Int32 offs = startIndex + (littleEndian ? index : lastByte - index);
data[offs] = (Byte) (value >> (8*index) & 0xFF);
}
}
The only requirement here is that you have to cast the input arg to 64-bit unsigned integer when passing it to the function.
public static unsafe int Reverse(int value)
{
byte* p = (byte*)&value;
return (*p << 24) | (p[1] << 16) | (p[2] << 8) | p[3];
}
If unsafe is allowed... Based on Marc Gravell's post
This will reverse the data inline if unsafe code is allowed...
fixed (byte* wavepointer = waveData)
new Span<byte>(wavepointer + offset, 4).Reverse();
If i take a uint value = 2921803 (0x2C954B), which is really a 4 byte package (4B 95 2C 00)
and i want to get the 16 least significant bits of the byte version of it using bitarray, how would i go about it?
This is how i am trying to do it:
byte[] bytes = BitConverter.GetBytes(value); //4B 95 2C 00 - bytes are moved around
BitArray bitArray = new BitArray(bytes); //entry [0] shows value for 1101 0010 (bits are reversed)
At this point, i am all turned around. I did try this:
byte[] bytes = BitConverter.GetBytes(value);
Array.Reverse(bytes);
BitArray bitArray = new BitArray(bytes);
Which gave me all the bits but completely reversed, reading from [31] to [0].
ultimately, i'm expecting/hoping to get 19349 (4B 95) as my answer.
This is how i was hoping to implement the function:
private uint GetValue(uint value, int bitsToGrab, int bitsToMoveOver)
{
byte[] bytes = BitConverter.GetBytes(value);
BitArray bitArray = new BitArray(bytes);
uint outputMask = (uint)(1 << (bitsToGrab - 1));
//now that i have all the bits, i can offset, and grab the ones i want
for (int i = bitsToMoveOver; i < bitsToGrab; i++)
{
if ((Convert.ToByte(bitArray[i]) & 1) > 0)
{
outputVal |= outputMask;
}
outputMask >>= 1;
}
}
The 16 least significant bits of 0x2C954B are 0x954B. You can get that as follows:
int value = 0x2C954B;
int result = value & 0xFFFF;
// result == 0x954B
If you want 0x4B95 then you can get that as follows:
int result = ((value & 0xFF) << 8) | ((value >> 8) & 0xFF);
// result == 0x4B95
Try this:
uint value = 0x002C954Bu;
int reversed = Reverse((int)value);
// reversed == 0x4B952C00;
int result = Extract(reversed, 16, 16);
// result == 0x4B95
with
int Extract(int value, int offset, int length)
{
return (value >> offset) & ((1 << length) - 1);
}
int Reverse(int value)
{
return ((value >> 24) & 0xFF) | ((value >> 8) & 0xFF00) |
((value & 0xFF00) << 8) | ((value & 0xFF) << 24);
}
unit - 32bits
Basically you should set 16 most significant bits to zero, so use bitwise AND operator:
uint newValue = 0x0000FFFF & uintValue;
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));
}
}