如何用 C# 计算 integer 数字的二进制表示中的尾随零

Question

如何有效地计算 integer 数字的二进制表示中尾随零的数量？

Answer 1

这是一个不错的、快速且简单的实现：

public static int NumberOfTrailingZeros(int i)
{
    return _lookup[(i & -i) % 37];
}

private static readonly int[] _lookup =
    {
        32, 0, 1, 26, 2, 23, 27, 0, 3, 16, 24, 30, 28, 11, 0, 13, 4, 7, 17,
        0, 25, 22, 31, 15, 29, 10, 12, 6, 0, 21, 14, 9, 5, 20, 8, 19, 18
    };

（取自http://graphics.stanford.edu/~seander/bithacks.html#ZerosOnRightModLookup 。）

Answer 2

只需从第一个数字开始制作一个掩码并继续移动它直到找到一些东西：

public static int numTrailingBinaryZeros(int n)
{
    int mask = 1;
    for (int i = 0; i < 32; i++, mask <<= 1)
        if ((n & mask) != 0)
            return i;

    return 32;
}

Answer 3

在java中它是这样实现的：

 public static int numberOfTrailingZeros(int i) { // HD, Figure 5-14 int y; if (i == 0) return 32; int n = 31; y = i <<16; if (y != 0) { n = n -16; i = y; } y = i << 8; if (y != 0) { n = n - 8; i = y; } y = i << 4; if (y != 0) { n = n - 4; i = y; } y = i << 2; if (y != 0) { n = n - 2; i = y; } return n - ((i << 1) >>> 31); }

我认为想法来自“黑客的喜悦”

Answer 4

int trailingZeros(int n) {
    if(!n)
        return 32;

    for(int s = 0; !(n & 1); s++)
        n >>= 1;
    return s;
}

Answer 5

Java 版本的这种变体使用了更多的一些技巧来删除最后一个条件测试：

    public static uint Ctz(uint num)
    {
        if (num == 0) return 32;    // optional, otherwise returns 0

        uint tmp;
        uint res = 0;
        num &= (uint)-(int)num;  // Isolate bit
        tmp = num >> 16; if (tmp != 0) { num = tmp; res += 16; }
        tmp = num >> 8;  if (tmp != 0) { num = tmp; res += 8; }
        tmp = num >> 4;  if (tmp != 0) { num = tmp; res += 4; }
        return res + ((num >> 1) - (num >> 3));
    }

Answer 6

Starting from.Net Core 3.0 there is BitOperations.TrailingZeroCount i dont know past versions but if i look into its current .net 7 source code it is inlined into Bmi1.TrailingZeroCount which is compiler intrinsic and converted to tzcnt x86 asm instruction on supported x86 environments. 它具有针对 ARM 的硬件加速实现，并作为最后的手段回退到软件实现。