I know this has been asked in similar variances many times, but I'm having trouble with the output of bitwise operations in C#(Unity3D).
I'm trying to do bit-reversal permutation, that is, get the bit-reversed order of integers(or unsigned int, either one) for the purpose using in the Cooley-Tukey FFT algorithm. So if I have 0, 1, 2, 3 - I want to end up with 0, 2, 1, 3 and if I have 0, 1, 2, 3, 4, 5, 6, 7 - I should get 0, 4, 2, 6, 1, 5, 3, 7.
I've tried a few bit-reversal algorithms found online, such as this one:
public uint ReverseBits(uint n)
{
n = (n >> 1) & 0x55555555 | (n << 1) & 0xaaaaaaaa;
n = (n >> 2) & 0x33333333 | (n << 2) & 0xcccccccc;
n = (n >> 4) & 0x0f0f0f0f | (n << 4) & 0xf0f0f0f0;
n = (n >> 8) & 0x00ff00ff | (n << 8) & 0xff00ff00;
n = (n >> 16) & 0x0000ffff | (n << 16) & 0xffff0000;
return n;
}
And I'd use it like this:
uint x = 1;
x = ReverseBits(x); //this results in x = 2147483648;
I wanted to try another algorithm so I found this one, which as pointed out reverses the bytes:
public uint ReverseBytes(uint value)
{
return (value & 0x000000FFU) << 24 | (value & 0x0000FF00U) << 8 |
(value & 0x00FF0000U) >> 8 | (value & 0xFF000000U) >> 24;
}
and I get the exact same number, x = 2147483648
. A bitwise operator such as >>
performs the same function in C# as it would in other languages such as C, right? So, am I missing a step?
The algorithms you are currently using reverse the bits in the whole integer (i.e. 32 bits for an int
and 64 bits for a long
), whereas what you really want is to reverse only the first k
bits (where n = 2^k
for the bit-reversal permutation).
A simple solution would be to use strings:
int x = 6;
int k = 3;
// Binary representation of x of length k
string binaryString = Convert.ToString(x, 2).PadLeft(k, '0');
int reversed = Convert.ToInt32(Reverse(binaryString), 2);
where Reverse
is defined as follows:
public static string Reverse( string s )
{
char[] charArray = s.ToCharArray();
Array.Reverse( charArray );
return new string( charArray );
}
Or if you don't want to use strings, you could stick with a bitwise operator solution:
int x = 6;
int k = 3;
int reversed = 0;
for(int i = 0; i < k; i++) {
// If the ith bit of x is toggled, toggle the ith bit from the right of reversed
reversed |= (x & (1 << i)) != 0 ? 1 << (k - 1 - i) : 0;
}
You can even remove the ternary operator at the cost of readability:
reversed |= (((x & (1 << i)) >> i) & 1) << (k - 1 - i);
The & 1
compensates for the case when the right arithmetic shift (>> i
) fills in the sign bit.
If you want to implement functions for given bit lengths (which you will if you know that your DFT has a given length such as 64) then you can hardcode various constants and write a function tailored to that bit-length, e.g:
public static int Reverse6Bits(int n)
{
n = (n >> 1) & 0x55 | (n << 1) & 0xaa;
n = (n >> 2) & 0x33 | (n << 2) & 0xcc;
n = (n >> 6) & 0x03 | (n << 2) & 0x3c;
return n;
}
if I have 0, 1, 2, 3, 4, 5, 6, 7 - I should get 0, 4, 2, 6, 1, 5, 3, 7
You can reverse 3 bits using a constant as a lookup table:
public static int Reverse3Bits(int n)
{
return (0x73516240 >> (n << 2)) & 7;
}
uint ret=n;
ret = ret >> 16 | ret<<16;
ret = (ret & 0xff00ff00) >> 8 | (ret & 0x00ff00ff) << 8;
ret = (ret & 0xf0f0f0f0) >> 4 | (ret & 0x0f0f0f0f) << 4;
ret = (ret & 0xcccccccc) >> 2 | (ret & 0x33333333) << 2;
ret = (ret & 0xaaaaaaaa) >> 1 | (ret & 0x55555555) << 1;
return ret;
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