c++ - boost::dynamic_bitset 的逆序

Question

有没有一种干净的方法来返回boost::dynamic_bitset对象的反向排序？

例如： 01001100 变成 00110010. 我能想到的最简单的解决方案是将位集转换为字符串，反转字符串并将其转换回位集，但这似乎是一种相当慢的方法，会使位串操作的速度无效。

先感谢您！

score 4 · Accepted Answer

boost::dynamic_bitset没有迭代器，所以一长串舒适的 STL 解决方案，例如，在我的脑海中，std::reverse或者std::swap它们的boost对应物不可用，我认为一个好方法是制作你自己的微不足道的反向方法：

#include <iostream>
#include <boost/dynamic_bitset.hpp>

void reverse(boost::dynamic_bitset<> &bs)
{
    for (size_t begin = 0, end = bs.size() - 1; begin < end; begin++, end--)
    {
        bool b = bs[end];
        bs[end] = bs[begin];
        bs[begin] = b;
    }
}

int main()
{
    size_t size = 8;
    boost::dynamic_bitset<> bs(size, 50);

    std::cout << "Normal:  " << bs << std::endl;
    reverse(bs);
    std::cout << "Reverse: " << bs << std::endl;
}

输出：

Normal:  00110010
Reverse: 01001100

现场演示

score 2 · Accepted Answer

你可以和非常幸运的人一起做得更好

 #define BOOST_DYNAMIC_BITSET_DONT_USE_FRIENDS

我首先注意到它是因为它在其他第三方库中使用（我忘记了名字，但它与 AI/ML 相关）。

我在这里有一个不是很通用的版本，因为它使用了特定于大小的位旋转技巧（反转例如字节或 uint32）。你可能对这些感兴趣：

您仍然可以看到 uint32 专用版本 Live On Compiler Explorer。

通用版

从那时起，我发现了这个不错的答案：在 C/C++ 中，反转字节中位顺序的最简单方法是什么？它为 2 的幂宽度整数类型提供了一种相当有效的就地反向算法。所以，现在我们有了完全通用的：

// make sure it's globally defined
#define BOOST_DYNAMIC_BITSET_DONT_USE_FRIENDS
#include <boost/dynamic_bitset.hpp>
#include <iostream>

template <typename Block, typename Allocator>
void reverse(boost::dynamic_bitset<Block, Allocator>& bs) {
    auto constexpr BLOCK_BIT = sizeof(Block) * CHAR_BIT;
    auto original_size       = bs.size();

    if (auto partial_block = bs.size() % BLOCK_BIT) {
        auto pad = (BLOCK_BIT - partial_block);
        bs.resize(bs.size() + pad);
        bs <<= pad;
    }

    // see https://stackoverflow.com/a/61109975/85371
    auto inplace = [](Block& n) {
        static_assert(std::is_unsigned_v<Block>);

        short bits = sizeof(n) * 8;
        Block mask = ~Block(0); // equivalent to uint32_t mask =
                                // 0b11111111111111111111111111111111;

        while (bits >>= 1) {
            mask ^= mask << (bits); // will convert mask to
                                    // 0b00000000000000001111111111111111;
            n = (n & ~mask) >> bits | (n & mask) << bits; // divide and conquer
        }
    };

    for (auto& b : bs.m_bits) {
        inplace(b);
    }
    std::reverse(begin(bs.m_bits), end(bs.m_bits));

    bs.resize(original_size);
}

请注意size()，对于BLOCK_BIT. 在某些情况下，这甚至可能导致人们更喜欢Block = std::uint8_t.

通用测试仪/基准测试

让我们写一些随机测试。为了便于实现，我们将反转的字符串表示与反转的字符串表示进行比较。

我添加了针对不同块大小的测试以及大小和时间的统计信息：

Live On 编译器资源管理器

// For quick testing
#include <random>
#include <chrono>
#include <boost/range/algorithm/reverse.hpp>
#include <boost/lexical_cast.hpp>
static auto now = std::chrono::high_resolution_clock::now;
using namespace std::chrono_literals;
#include <boost/accumulators/accumulators.hpp>
#include <boost/accumulators/statistics.hpp>

namespace ba  = boost::accumulators;
namespace bat = ba::tag;

template <typename Block> bool run_test(unsigned n, auto& stats) {
    using BitSet = boost::dynamic_bitset<Block>;

    auto gen = std::bind(std::uniform_int_distribution<Block>{},
                         std::mt19937{std::random_device{}()});

    while (n--) {
        Block init[]{gen(), gen(), gen()};
        auto sz = std::bind(
                std::uniform_int_distribution(0ul, sizeof(init) * CHAR_BIT),
                std::mt19937{std::random_device{}()});

        BitSet bs(std::begin(init), std::end(init));
        bs.resize(sz());
        stats(bs.size());

        std::string expected = boost::lexical_cast<std::string>(bs);
        boost::reverse(expected);

        BitSet revclone = bs;
        reverse(revclone);

        auto actual = boost::lexical_cast<std::string>(revclone);
        if (expected != actual) {
            std::cout << __PRETTY_FUNCTION__ << " '" << bs << "': \n"
                      << "  expected: " << expected << "\n"
                      << "  actual: " << actual << "\n";
            return false;
        }
    }
    return true;
}

int main() {
    auto start = now();
    ba::accumulator_set<double,
                        ba::stats<bat::mean, bat::variance, bat::min, bat::max>>
        stats;

    if (run_test<unsigned char>(10'000, stats))
        std::cout << "Completed 10'000 tests with char blocks\n";
    if (run_test<uint16_t>(10'000, stats))
        std::cout << "Completed 10'000 tests with uint16_t blocks\n";
    if (run_test<uint32_t>(100'000, stats))
        std::cout << "Completed 100'000 tests with uint32_t blocks\n";
    if (run_test<uintmax_t>(1'000'000, stats))
        std::cout << "Completed 1'000'000 tests with uintmax_t blocks\n";

    auto cost = ((now() - start)/1.us) / ba::count(stats);

    std::cout
        << "Samples " << ba::count(stats)
        << " mean: " << ba::mean(stats)
        << " min: " << ba::min(stats)
        << " max: " << ba::max(stats)
        << " stddev: " << std::sqrt(ba::variance(stats))
        << "\n";

    std::cout << "Average cost " << cost << "μs\n";
}

在我的机器上打印：

Completed 10'000 tests with char blocks
Completed 10'000 tests with uint16_t blocks
Completed 100'000 tests with uint32_t blocks
Completed 1'000'000 tests with uintmax_t blocks
Samples 1120000 mean: 90.3283 min: 0 max: 192 stddev: 55.9233
Average cost 3.69335μs

real    0m4,141s
user    0m4,061s
sys     0m0,003s

~~因此，在平均大小为 90 位的情况下，最多 192 位的位集可以在不到 4μs 的时间内反转。不错。~~

适当的微观基准

使用Nonius，我们可以从可预测的测试中获得可靠的数据。对于 31、32、37 位大小，净时序在 10-30ns 范围内。

使用的代码：

#define NONIUS_RUNNER
#include <nonius/nonius.h++>
#include <nonius/main.h++>

template <typename Block> void run_test(nonius::chronometer& cm, size_t target_size) {
    using BitSet = boost::dynamic_bitset<Block>;

    static const std::string data{
        "0100110111010010010001100111010010010001011100100100111010100010011010"
        "01100000011000010001110111"};

    BitSet bs(data, 0, target_size);
    assert(bs.size() == target_size);

    cm.measure([&] { reverse(bs); });
}

NONIUS_BENCHMARK("Block=uchar,     sz=32", [](nonius::chronometer cm) { run_test<uint8_t>(cm,   32); })
NONIUS_BENCHMARK("Block=uint16_t,  sz=32", [](nonius::chronometer cm) { run_test<uint16_t>(cm,  32); })
NONIUS_BENCHMARK("Block=uint32_t,  sz=32", [](nonius::chronometer cm) { run_test<uint32_t>(cm,  32); })
NONIUS_BENCHMARK("Block=uintmax_t, sz=32", [](nonius::chronometer cm) { run_test<uintmax_t>(cm, 32); })

NONIUS_BENCHMARK("Block=uchar,     sz=31", [](nonius::chronometer cm) { run_test<uint8_t>(cm,   31); })
NONIUS_BENCHMARK("Block=uint16_t,  sz=31", [](nonius::chronometer cm) { run_test<uint16_t>(cm,  31); })
NONIUS_BENCHMARK("Block=uint32_t,  sz=31", [](nonius::chronometer cm) { run_test<uint32_t>(cm,  31); })
NONIUS_BENCHMARK("Block=uintmax_t, sz=31", [](nonius::chronometer cm) { run_test<uintmax_t>(cm, 31); })

NONIUS_BENCHMARK("Block=uchar,     sz=37", [](nonius::chronometer cm) { run_test<uint8_t>(cm,   37); })
NONIUS_BENCHMARK("Block=uint16_t,  sz=37", [](nonius::chronometer cm) { run_test<uint16_t>(cm,  37); })
NONIUS_BENCHMARK("Block=uint32_t,  sz=37", [](nonius::chronometer cm) { run_test<uint32_t>(cm,  37); })
NONIUS_BENCHMARK("Block=uintmax_t, sz=37", [](nonius::chronometer cm) { run_test<uintmax_t>(cm, 37); })

完整的交互式图表：http ://stackoverflow-sehe.s3.amazonaws.com/974d10e8-74ae-4fcf-be03-6a0d0e01b5ad/stats.html

c++ - boost::dynamic_bitset 的逆序

2 回答 2

通用版

通用测试仪/基准测试

适当的微观基准

Related

Reference