从UInt64 Number的startPos开始，获取n位的效率如何

How effective get n bits, starting from startPos from the UInt64 Number.

从UInt64 Number的startPos开始，获取n位的效率如何。

i know woh get bit by bit, but i want to do in more effective way.

我知道我会一点一点地得到，但我想以更有效的方式做。

    public static ulong GetBits(ulong value, int startPos)
    {
        int mask = 1 << startPos;
        ulong masked_n = value & (ulong)mask;
        ulong thebit = masked_n >> startPos;
        return (ulong)thebit;
    }

2 个解决方案

#1

// assuming bit numbers start with 0, and that
// startPos is the position of the desired
// least-significant (lowest numbered) bit

public static ulong GetBits( ulong value, int startPos, int bits )
{
    ulong mask = ( ( 1UL << bits ) - 1 ) << startPos;
    return ( value & mask ) >> startPos;
}

#2

Ok - so let's say (for sanity's sake, let's talk 8-bits) you have:

好的 - 所以让我们说（为了理智，让我们谈谈8位）你有：

10101010

And you want 3 (m) bits starting at bit 2 (n). You you'll need a mask like this:

你想要从第2位（n）开始的3（m）位。你需要这样的面具：

source:  10101010
mask:    00011100
&result: 00001000

So how to generate the mask? We start with 1 and shift it by the number of bits we want (m)

那么如何生成面具？我们从1开始并将它移动我们想要的位数（m）

start:        00000001
start << 3:   00001000

Now we need a three 1's in our mask, so we simply minus one from the last step:

现在我们需要在面具中使用三个1，所以我们在最后一步中减去一个：

00001000 - 1 = 00000111

So we almost have our mask, now we just need to line it up by shifting it by 2 (n)

所以我们几乎有了我们的面具，现在我们只需要通过将它移动2（n）来排列它

00000111 << 2 = 00011100

And we have our answer

我们有答案

#1

// assuming bit numbers start with 0, and that
// startPos is the position of the desired
// least-significant (lowest numbered) bit

public static ulong GetBits( ulong value, int startPos, int bits )
{
    ulong mask = ( ( 1UL << bits ) - 1 ) << startPos;
    return ( value & mask ) >> startPos;
}

#2