如何解决boost::dynamic_bitset 的倒序
是否有一种干净的方法可以返回 boost::dynamic_bitset
对象的相反顺序?
例如:
01001100
变成
00110010
。
我能想到的最简单的解决方案是将 bitset 转换为字符串,反转字符串并将其转换回 bitset,但这似乎是一种相当慢的方法,它使 bitstring 操作的速度无效。
先谢谢你!
解决方法
boost::dynamic_bitset
没有迭代器,所以有很多舒适的 STL 解决方案,比如我想不到的 std::reverse
或 std::swap
或它们的 boost
对应物不可用,我认为最好的方法是制作自己的简单反向方法:
#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
,
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(我忘记了名称,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操作技巧(反转例如字节或 uint32)。您可能对这些感兴趣:
- 见http://aggregate.org/MAGIC/#Bit%20Reversal
- 或https://graphics.stanford.edu/~seander/bithacks.html#ReverseParallel
您仍然可以看到 uint32 专用版本Live On Compiler Explorer。
通用版本
从那时起,我发现了这个不错的答案:In C/C++ what's the simplest way to reverse the order of bits in a byte?,它为 2 次幂宽度整数类型提供了一个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);
}
注意,对于 size()
的 BLOCK_BIT
倍数,反转会更有效。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
让我们编写一些随机测试。为了便于实现,我们将反向字符串表示与反向字符串表示进行比较。
我添加了不同块大小的测试以及大小和时间的统计数据:
// 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";
}
打印,在我的机器上:
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,平均大小为 90 位,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,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,[](nonius::chronometer cm) { run_test<uint16_t>(cm,32); })
NONIUS_BENCHMARK("Block=uint32_t,[](nonius::chronometer cm) { run_test<uint32_t>(cm,32); })
NONIUS_BENCHMARK("Block=uintmax_t,[](nonius::chronometer cm) { run_test<uintmax_t>(cm,32); })
NONIUS_BENCHMARK("Block=uchar,sz=31",31); })
NONIUS_BENCHMARK("Block=uint16_t,31); })
NONIUS_BENCHMARK("Block=uint32_t,31); })
NONIUS_BENCHMARK("Block=uintmax_t,31); })
NONIUS_BENCHMARK("Block=uchar,sz=37",37); })
NONIUS_BENCHMARK("Block=uint16_t,37); })
NONIUS_BENCHMARK("Block=uint32_t,37); })
NONIUS_BENCHMARK("Block=uintmax_t,37); })
完整的交互式图表:http://stackoverflow-sehe.s3.amazonaws.com/974d10e8-74ae-4fcf-be03-6a0d0e01b5ad/stats.html
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