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fast way to convert integer array to byte array (11 bit)

I have integer array and I need to convert it to byte array
but I need to take (only and just only) first 11 bit of each element of the هinteger array
and then convert it to a byte array
I tried this code
// ***********convert integer values to byte values
//***********to avoid the left zero padding on the byte array
// *********** first step : convert to binary string
// ***********second step : convert binary string to byte array
// *********** first step
string ByteString = Convert.ToString(IntArray[0], 2).PadLeft(11,'0');
for (int i = 1; i < IntArray.Length; i++)
ByteString = ByteString + Convert.ToString(IntArray[i], 2).PadLeft(11, '0');
// ***********second step
int numOfBytes = ByteString.Length / 8;
byte[] bytes = new byte[numOfBytes];
for (int i = 0; i < numOfBytes; ++i)
{
bytes[i] = Convert.ToByte(ByteString.Substring(8 * i, 8), 2);
}
But it takes too long time (if the file size large , the code takes more than 1 minute)
I need a very very fast code (very few milliseconds only )
can any one help me ?

Basically, you're going to be doing a lot of shifting and masking. The exact nature of that depends on the layout you want. If we assume that we pack little-endian from each int, appending on the left, so two 11-bit integers with positions:
abcdefghijk lmnopqrstuv
become the 8-bit chunks:
defghijk rstuvabc 00lmnopq
(i.e. take the lowest 8 bits of the first integer, which leaves 3 left over, so pack those into the low 3 bits of the next byte, then take the lowest 5 bits of the second integer, then finally the remaining 6 bits, padding with zero), then something like this should work:
using System;
using System.Linq;
static class Program
{
static string AsBinary(int val) => Convert.ToString(val, 2).PadLeft(11, '0');
static string AsBinary(byte val) => Convert.ToString(val, 2).PadLeft(8, '0');
static void Main()
{
int[] source = new int[1432];
var rand = new Random(123456);
for (int i = 0; i < source.Length; i++)
source[i] = rand.Next(0, 2047); // 11 bits
// Console.WriteLine(string.Join(" ", source.Take(5).Select(AsBinary)));
var raw = Encode(source);
// Console.WriteLine(string.Join(" ", raw.Take(6).Select(AsBinary)));
var clone = Decode(raw);
// now prove that it worked OK
if (source.Length != clone.Length)
{
Console.WriteLine($"Length: {source.Length} vs {clone.Length}");
}
else
{
int failCount = 0;
for (int i = 0; i < source.Length; i++)
{
if (source[i] != clone[i] && failCount++ == 0)
{
Console.WriteLine($"{i}: {source[i]} vs {clone[i]}");
}
}
Console.WriteLine($"Errors: {failCount}");
}
}
static byte[] Encode(int[] source)
{
long bits = source.Length * 11;
int len = (int)(bits / 8);
if ((bits % 8) != 0) len++;
byte[] arr = new byte[len];
int bitOffset = 0, index = 0;
for (int i = 0; i < source.Length; i++)
{
// note: this encodes little-endian
int val = source[i] & 2047;
int bitsLeft = 11;
if(bitOffset != 0)
{
val = val << bitOffset;
arr[index++] |= (byte)val;
bitsLeft -= (8 - bitOffset);
val >>= 8;
}
if(bitsLeft >= 8)
{
arr[index++] = (byte)val;
bitsLeft -= 8;
val >>= 8;
}
if(bitsLeft != 0)
{
arr[index] = (byte)val;
}
bitOffset = bitsLeft;
}
return arr;
}
private static int[] Decode(byte[] source)
{
int bits = source.Length * 8;
int len = (int)(bits / 11);
// note no need to worry about remaining chunks - no ambiguity since 11 > 8
int[] arr = new int[len];
int bitOffset = 0, index = 0;
for(int i = 0; i < source.Length; i++)
{
int val = source[i] << bitOffset;
int bitsLeftInVal = 11 - bitOffset;
if(bitsLeftInVal > 8)
{
arr[index] |= val;
bitOffset += 8;
}
else if(bitsLeftInVal == 8)
{
arr[index++] |= val;
bitOffset = 0;
}
else
{
arr[index++] |= (val & 2047);
if(index != arr.Length) arr[index] = val >> 11;
bitOffset = 8 - bitsLeftInVal;
}
}
return arr;
}
}
If you need a different layout you'll need to tweak it.
This encodes 512 MiB in just over a second on my machine.
Overview to the Encode method:
The first thing is does is pre-calculate the amount of space that is going to be required, and allocate the output buffer; since each input contributes 11 bits to the output, this is just some modulo math:
long bits = source.Length * 11;
int len = (int)(bits / 8);
if ((bits % 8) != 0) len++;
byte[] arr = new byte[len];
We know the output position won't match the input, and we know we're going to be starting each 11-bit chunk at different positions in bytes each time, so allocate variables for those, and loop over the input:
int bitOffset = 0, index = 0;
for (int i = 0; i < source.Length; i++)
{
...
}
return arr;
So: taking each input in turn (where the input is the value at position i), take the low 11 bits of the value - and observe that we have 11 bits (of this value) still to write:
int val = source[i] & 2047;
int bitsLeft = 11;
Now, if the current output value is partially written (i.e. bitOffset != 0), we should deal with that first. The amount of space left in the current output is 8 - bitOffset. Since we always have 11 input bits we don't need to worry about having more space than values to fill, so: left-shift our value by bitOffset (pads on the right with bitOffset zeros, as a binary operation), and "or" the lowest 8 bits of this with the output byte. Essentially this says "if bitOffset is 3, write the 5 low bits of val into the 5 high bits of the output buffer"; finally, fixup the values: increment our write position, record that we have fewer bits of the current value still to write, and use right-shift to discard the 8 low bits of val (which is made of bitOffset zeros and 8 - bitOffset "real" bits):
if(bitOffset != 0)
{
val = val << bitOffset;
arr[index++] |= (byte)val;
bitsLeft -= (8 - bitOffset);
val >>= 8;
}
The next question is: do we have (at least) an entire byte of data left? We might not, if bitOffset was 1 for example (so we'll have written 7 bits already, leaving just 4). If we do, we can just stamp that down and increment the write position - then once again track how many are left and throw away the low 8 bits:
if(bitsLeft >= 8)
{
arr[index++] = (byte)val;
bitsLeft -= 8;
val >>= 8;
}
And it is possible that we've still got some left-over; for example, if bitOffset was 7 we'll have written 1 bit in the first chunk, 8 bits in the second, leaving 2 more to write - or if bitOffset was 0 we won't have written anything in the first chunk, 8 in the second, leaving 3 left to write. So, stamp down whatever is left, but do not increment the write position - we've written to the low bits, but the next value might need to write to the high bits. Finally, update bitOffset to be however many low bits we wrote in the last step (which could be zero):
if(bitsLeft != 0)
{
arr[index] = (byte)val;
}
bitOffset = bitsLeft;
The Decode operation is the reverse of this logic - again, calculate the sizes and prepare the state:
int bits = source.Length * 8;
int len = (int)(bits / 11);
int[] arr = new int[len];
int bitOffset = 0, index = 0;
Now loop over the input:
for(int i = 0; i < source.Length; i++)
{
...
}
return arr;
Now, bitOffset is the start position that we want to write to in the current 11-bit value, so if we start at the start, it will be 0 on the first byte, then 8; 3 bits of the second byte join with the first 11-bit integer, so the 5 bits become part of the second - so bitOffset is 5 on the 3rd byte, etc. We can calculate the number of bits left in the current integer by subtracting from 11:
int val = source[i] << bitOffset;
int bitsLeftInVal = 11 - bitOffset;
Now we have 3 possible scenarios:
1) if we have more than 8 bits left in the current value, we can stamp down our input (as a bitwise "or") but do not increment the write position (as we have more to write for this value), and note that we're 8-bits further along:
if(bitsLeftInVal > 8)
{
arr[index] |= val;
bitOffset += 8;
}
2) if we have exactly 8 bits left in the current value, we can stamp down our input (as a bitwise "or") and increment the write position; the next loop can start at zero:
else if(bitsLeftInVal == 8)
{
arr[index++] |= val;
bitOffset = 0;
}
3) otherwise, we have less than 8 bits left in the current value; so we need to write the first bitsLeftInVal bits to the current output position (incrementing the output position), and whatever is left to the next output position. Since we already left-shifted by bitOffset, what this really means is simply: stamp down (as a bitwise "or") the low 11 bits (val & 2047) to the current position, and whatever is left (val >> 11) to the next if that wouldn't exceed our output buffer (padding zeros). Then calculate our new bitOffset:
else
{
arr[index++] |= (val & 2047);
if(index != arr.Length) arr[index] = val >> 11;
bitOffset = 8 - bitsLeftInVal;
}
And that's basically it. Lots of bitwise operations - shifts (<< / >>), masks (&) and combinations (|).

If you wanted to store the least significant 11 bits of an int into two bytes such that the least significant byte has bits 1-8 inclusive and the most significant byte has 9-11:
int toStore = 123456789;
byte msb = (byte) ((toStore >> 8) & 7); //or 0b111
byte lsb = (byte) (toStore & 255); //or 0b11111111
To check this, 123456789 in binary is:
0b111010110111100110100010101
MMMLLLLLLLL
The bits above L are lsb, and have a value of 21, above M are msb and have a value of 5
Doing the work is the shift operator >> where all the binary digits are slid to the right 8 places (8 of them disappear from the right hand side - they're gone, into oblivion):
0b111010110111100110100010101 >> 8 =
0b1110101101111001101
And the mask operator & (the mask operator works by only keeping bits where, in each position, they're 1 in the value and also 1 in the mask) :
0b111010110111100110100010101 &
0b000000000000000000011111111 (255) =
0b000000000000000000000010101
If you're processing an int array, just do this in a loop:
byte[] bs = new byte[ intarray.Length*2 ];
for(int x = 0, b=0; x < intarray.Length; x++){
int toStore = intarray[x];
bs[b++] = (byte) ((toStore >> 8) & 7);
bs[b++] = (byte) (toStore & 255);
}

Get, Set, Read, converts bitwise in Short type value in C#

I have a short value X:
short X=1; //Result in binary: 0000000000000001
I need to split them into an array and set the bits (say bit 6 and 10) //Result in binary: 0000001000100001
I need to convert it back to short X value.
How can I do it painlessly?
Could you please help?

1. Manual solution
Setting bit 6 and 10:
myValue |= (1 << 6)|(1 << 10);
Clearing bit 6 and 10:
myValue &= ~((1 << 6)|(1 << 10));
2. Use BitArray
var bits = new BitArray(16); // 16 bits
bits[5] = true;
bits[10] = true;
Convert back to short:
var raw = new byte[2];
bits.CopyTo(raw, 0);
var asShort = BitConverter.ToInt16(raw, 0);

If what you are referring to is a very basic encryption, then perhaps using the XOR (^) operator would be better suited for your needs.
short FlipBytes(short original, params int[] bytesToSet)
{
int key = 0;
foreach (int b in bytesToSet)
{
if (b >= 0 && b < 16)
{
key |= 1 << b;
}
}
return (short)(original ^ key);
}
This method will both set and reset the bytes that you desire. For example:
short X = 1;
short XEncrypt = FlipBytes(X, 6, 10);
short XDecrypt = FlipBytes(XEncrypt, 6, 10);
// X = 1 , Binary = 0000000000000001
// XEncrypt = 1089 , Binary = 0000010001000001
// XDecrypt = 1 , Binary = 0000000000000001

If you have a int value "intValue" and you want to set a specific bit at position "bitPosition", do something like:
intValue = intValue | (1 << bitPosition);
or shorter:
intValue |= 1 << bitPosition;
If you want to reset a bit (i.e, set it to zero), you can do this:
intValue &= ~(1 << bitPosition);
(The operator ~ reverses each bit in a value, thus ~(1 << bitPosition) will result in an int where every bit is 1 except the bit at the given bitPosition.)

Linq solution, terser, but probably, less readable than foreach loop:
using System.Linq;
...
short X = 1;
var bitsToSet = new[] { 5, 9 };
var result = X | bitsToSet.Aggregate((s, a) => s |= 1 << a);
If you insist on short add cast:
short result = (short) (X | bitsToSet.Aggregate((s, a) => s |= 1 << a));

How to merge two bitmaps with specific shift(offset)?

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 ) ?

Convert int to little-endian formated bytes in C++ for blobId in Azure

Working with a base64 encoding for Azure (http://msdn.microsoft.com/en-us/library/dd135726.aspx) and I dont seem to work out how to get the required string back. I'm able to do this in C# where I do the following.
int blockId = 5000;
var blockIdBytes = BitConverter.GetBytes(blockId);
Console.WriteLine(blockIdBytes);
string blockIdBase64 = Convert.ToBase64String(blockIdBytes);
Console.WriteLine(blockIdBase64);
Which prints out (in LINQPad):
Byte[] (4 items)
| 136 |
| 19 |
| 0 |
| 0 |
iBMAAA==
In Qt/C++ I tried a few aporaches, all of them returning the wrong value.
const int a = 5000;
QByteArray b;
for(int i = 0; i != sizeof(a); ++i) {
b.append((char)(a&(0xFF << i) >>i));
}
qDebug() << b.toBase64(); // "iIiIiA=="
qDebug() << QByteArray::number(a).toBase64(); // "NTAwMA=="
qDebug() << QString::number(a).toUtf8().toBase64(); // "NTAwMA=="
How can I get the same result as the C# version?

See my comment for the problem with your for loop. It's shifting by one bit more each pass, but actually it should be 8 bits. Personally, I prefer this to a loop:
b.append(static_cast<char>(a >> 24));
b.append(static_cast<char>((a >> 16) & 0xff));
b.append(static_cast<char>((a >> 8) & 0xff));
b.append(static_cast<char>(a & 0xff));
The code above is for network standard byte order (big endian). Flip the order of the four operations from last to first for little endian byte order.

I ended up doing the following:
QByteArray temp;
int blockId = 5000;
for(int i = 0; i != sizeof(blockId); i++) {
temp.append((char)(blockId >> (i * 8)));
}
qDebug() << temp.toBase64(); // "iBMAAA==" which is correct

I think this would be clearer, though may be claimed to be ill styled...
int i = 0x01020304;
char (&bytes)[4] = (char (&)[4])i;
and you can access each byte directly with bytes[0], bytes[1], ... and do what ever you want to do with them.

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));
}
}