Bit twiddling：从ulong，获取表示哪些字节非零的位掩码

Question

Say we have an 8-byte ulong. For each byte, we want to know whether it is zero or non-zero. The desired result is a byte, whose 8 bits represent the "non-zeroness" of the original 8 bytes.

Is there a name for this operation or set of operations?

How can we achieve this very efficiently? An ideal solution would be branchless.

As an alternate requirement, a useful answer would be the position of the first non-zero byte. Eg if the first non-zero byte is the third one, the answer would be 2 (0-based index). I realize this can be approached by counting the leading zeros of the initial requirement's answer, but perhaps this will allow a shortcut.

Answer 1

This might help -

    private void Evaluate(ulong n)
    {
        ulong f = 255;
        int r = 0, p = -1;
        for (int i = 0; i < 8; i++)
        {
            r >>= 1;
            var t = n & f;
            if (t > 0)
            {
                r += 128;
                if (p < 0)
                    p = i;
            }
            f <<= 8;
        }
        Console.WriteLine($"Resulting byte: {r}");
        Console.WriteLine($"Position: {p}");
    }

What I have done is bitwise ANDed every byte of input 8 byte number with 255(1111 1111). If the result is one I right shifted one bit for result and added 128(1000 000).

For position I initialized p = -1 in case number is zero. Otherwise, assign first index of `> 0'.

There are lot of optimization possible like comparing input number to zero and if true simply return 0 for result and -1 for position.

Answer 2

You can shift the long by 8 bits and look at each byte independently (ie check if the byte == 0). Using the index of the byte, you can set the value in the resulting byte by shifting a 1 into that bit index.

    private byte TestULong(ulong value)
    {
        byte result = 0;
        for (int i = 0; i < 8; i++)
        {
            var test = (byte)(value >> (i * 8));
            if (test != 0)
            {
                result = (byte)(result | (1 << i));
            }
        }
        return result;
    }

Answer 3

I have a feeling that there is a purely math-based approach to this, but I can't seem to suss it out. Otherwise, the most efficient way to do this would be with a loop and some bit-wise comparisons:

public static byte Evaluate(ulong n) {
    ulong mask = 0xFF;
    byte result = 0;
    for (int i = 0; i < 8; i++) {
        if ((n & (mask << (i * 8))) != 0) {
            result |= (byte)(1 << i);
        }
    }
    return result;
}

The mask is a preconfigured value where every bit in a single byte is 1 (255 in decimal, FF in hexadecimal). You offset the mask by i * 8 to get it to cover the nth byte of the input and then use a bitwise-AND to get the value of that byte. All you need to do is check if that byte value is non-zero, and if it is, set the corresponding bit of the result byte to 1. (This can be done with either an addition or a bitwise-OR, so I opted to go the OR route to keep with the bitwise theme.)

https://dotnetfiddle.net/EA7NRw