将范围拆分成重叠范围的范围
我尝试使用Ranges-V3库将一个容器值切片到一系列范围内，以便相邻范围共享边界元素。
考虑以下：
using namespace ranges;
std::vector<int> v = { 1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9 };
auto myRanges = v | /* something like adjacent split */
for_each( myRanges, []( auto&& range ){ std::cout << range << std::endl;} );
我想根据该地区是否满足两个标准将范围划分为重叠的子范围：
期望的输出：
[1,2,3]
[3,0,4,0,5,0,6]
[6,7,8]
[8,0,0,9]
我的尝试：
auto degenerate =
[]( auto&& arg ){
return distance( arg ) < 2;
};
auto myRanges = v | view::split(0) | view::remove_if( degenerate );
for_each( myRanges, []( auto&& range ){ std::cout << range << std::endl;} );
输出：
[1,2,3]
[6,7,8]
我对自己的方式感到不知所措
如果我正确理解你的需求，你可以根据adjacent_find实现一个生成器：
namespace detail {
template<typename IterT, typename SentT>
struct seg_gen_fn {
IterT it_;
SentT end_;
bool parity_ = true;
ranges::iterator_range<IterT> operator ()() {
if (it_ == end_) {
return {it_, it_};
}
auto n = ranges::adjacent_find(
it_, end_,
[p = std::exchange(parity_, !parity_)](auto const a, auto const b) {
return a && !b == p;
}
);
return {
std::exchange(it_, n),
n != end_ ? ranges::next(std::move(n)) : std::move(n)
};
}
};
template<typename RngT>
constexpr auto seg_gen(RngT&& rng)
-> seg_gen_fn<ranges::iterator_t<RngT>, ranges::sentinel_t<RngT>>
{ return {ranges::begin(rng), ranges::end(rng)}; }
} // namespace detail
auto const segmented_view = [](auto&& rng) {
return ranges::view::generate(detail::seg_gen(decltype(rng)(rng)))
| ranges::view::take_while([](auto const& seg) { return !seg.empty(); });
};
int main() {
auto const ns = {1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9};
ranges::copy(segmented_view(ns), ranges::ostream_iterator<>{std::cout, "n"});
}
在线演示
并不完全像人们希望的那样简洁......： - [
这对于一次性代码来说可能是好的，但是可以做更多的工作，并且可以更加可重用：
namespace detail {
namespace tag = ranges::tag;
template<
typename RngT, typename PredT, typename IterT = ranges::iterator_t<RngT>,
typename StateT = ranges::tagged_compressed_tuple<
tag::begin(IterT), tag::end(ranges::sentinel_t<RngT>),
tag::current(bool), tag::fun(ranges::semiregular_t<PredT>)
>
>
struct seg_gen_fn : private StateT {
constexpr seg_gen_fn(RngT&& rng, PredT pred)
: StateT{ranges::begin(rng), ranges::end(rng), true, std::move(pred)}
{ }
ranges::iterator_range<IterT> operator ()() {
StateT& state = *this;
auto& it = state.begin();
if (it == state.end()) {
return {it, it};
}
auto& parity = state.current();
auto n = ranges::adjacent_find(
it, state.end(),
[p = std::exchange(parity, !parity), &pred = state.fun()]
(auto const& a, auto const& b) {
return !pred(a) && pred(b) == p;
}
);
return {
std::exchange(it, n),
n != state.end() ? ranges::next(std::move(n)) : std::move(n)
};
}
};
template<typename RngT, typename PredT>
constexpr seg_gen_fn<RngT, PredT> seg_gen(RngT&& rng, PredT pred) {
return {std::forward<RngT>(rng), std::move(pred)};
}
} // namespace detail
auto const segmented_view = [](auto&& rng, auto pred) {
return ranges::view::generate(detail::seg_gen(decltype(rng)(rng), std::move(pred)))
| ranges::view::take_while([](auto const& seg) { return !seg.empty(); });
};
int main() {
auto const ns = {1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9};
ranges::copy(
segmented_view(ns, [](auto const n) { return n == 0; }),
ranges::ostream_iterator<>{std::cout, "n"}
);
}
在线演示
概念检查和预测留作练习。
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