如何获得BitSet表示的位数？

Question

If I run the following code:

BitSet test = new BitSet(20);
System.out.println("Bitset size is " + test.size());
System.out.println("Bitset length is " + test.length());

I get the output:

Bitset size is 64
Bitset length is 0

Which makes sense now that I look closer at the documentation (the former getting the size including implementation overhead and the latter being the last set bit, with all defaulting to false), but is not what I want.

Since the BitSet I actually use can have varying lengths, I want to be able to get back the number of bits represented (ignoring whether they are set). Is there a way to do that (hopefully with a built-in method)?

I realize I could try flipping all the bits and then doing length, or by tracking the variable I use to instantiate the BitSet, but I feel like both would require several lines of comments with my reasons and I was hoping for something a bit more self-documenting.

Answer 1

A BitSet will automatically expand to a size sufficient to represent the highest bit set in it; the initial size is simply a hint as to how many bits you intend to use. So your question as posed makes no sense.

Answer 2

我在Mac上进行了一些测试，看来BitSet的长度希望是64的倍数。因此，对于您的BitSet大小为20，它默认为64。如果将大小设置为65，则应该看到一个长度设置为129。将其设置为129应该给您192。这应该给您一种计算“真实” BitSet长度的方法。

Answer 3

Bitset internally uses long (64 bits). So you can determine number of bits needed in multiples of this using the size() method.