从整数数组构造位集

Question

It's easy to construct a bitset<64> from a uint64_t :

uint64_t flags = ...;
std::bitset<64> bs{flags};

But is there a good way to construct a bitset<64 * N> from a uint64_t[N] , such that flags[0] would refer to the lowest 64 bits?

uint64_t flags[3];
// ... some assignments
std::bitset<192> bs{flags};  // this very unhelpfully compiles
                             // yet is totally invalid

Or am I stuck having to call set() in a loop?

Answer 1

std::bitset has no range constructor, so you will have to loop, but setting every bit individually with std::bitset::set() is underkill. std::bitset has support for binary operators, so you can at least set 64 bits in bulk:

  std::bitset<192> bs;

  for(int i = 2; i >= 0; --i) {
    bs <<= 64;
    bs |= flags[i];
  }

Update: In the comments, @icando raises the valid concern that bitshifts are O(N) operations for std::bitset s. For very large bitsets, this will ultimately eat the performance boost of bulk processing. In my benchmarks, the break-even point for a std::bitset<N * 64> in comparison to a simple loop that sets the bits individually and does not mutate the input data:

int pos = 0;
for(auto f : flags) {
  for(int b = 0; b < 64; ++b) {
    bs.set(pos++, f >> b & 1);
  }
}

is somewhere around N == 200 (gcc 4.9 on x86-64 with libstdc++ and -O2 ). Clang performs somewhat worse, breaking even around N == 160 . Gcc with -O3 pushes it up to N == 250 .

Taking the lower end, this means that if you want to work with bitsets of 10000 bits or larger, this approach may not be for you. On 32-bit platforms (such as common ARMs), the threshold will probably lie lower, so keep that in mind when you work with 5000-bit bitsets on such platforms. I would argue, however, that somewhere far before this point, you should have asked yourself if a bitset is really the right choice of container.

Answer 2

If initializing from range is important, you might consider using std::vector

It does have constructor from pair of iterators