范围修剪视图实现不适用于反向视图

Question

我写了一个名为trim的 C++20 范围视图(不是 range-v3 视图(，给定一个范围和一个一元谓词，返回一个新范围，没有满足谓词的前后元素。(range-v3 库有这样的视图，但在 C++20 中缺少它。

这是实现(这可能不是最佳的，但是关于范围库的文档并不多，所以鉴于资源有限，这是我能想到的(：

namespace rg = std::ranges;
// -------- trim ----------
template<rg::input_range R, typename P> requires rg::view<R>
class trim_view : public rg::view_interface<trim_view<R, P>>
{
private:
R base_ {};
P pred_;
mutable rg::iterator_t<R> iter_ {std::begin(base_)};
mutable rg::iterator_t<R> end_  {std::end(base_)};
public:
trim_view() = default;
constexpr trim_view(R base, P pred)
: base_(std::move(base)), pred_(std::move(pred)), iter_(std::begin(base_)), end_(std::end(base_))
{}
constexpr R base() const &
{return base_;}
constexpr R base() && 
{return std::move(base_);}
constexpr auto begin()
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() const requires rg::range<const R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() requires rg::random_access_range<R> && rg::sized_range<R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() const requires rg::random_access_range<const R> && rg::sized_range<const R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto end()
{ return end_ ; }
constexpr auto end() const requires rg::range<const R>
{ return end_ ; }
constexpr auto end() requires rg::random_access_range<R> && rg::sized_range<R>
{ return end_ ; }
constexpr auto end() const requires rg::random_access_range<const R> && rg::sized_range<const R>
{ return end_ ; }
constexpr auto size() requires rg::sized_range<R>
{ return std::distance(iter_, end_); }    
constexpr auto size() const requires rg::sized_range<const R>
{ return std::distance(iter_, end_); }
};
template<class R, typename P>
trim_view(R&& base, P pred)
-> trim_view<rg::views::all_t<R>, P>;
namespace details
{
template <typename P>
struct trim_view_range_adaptor_closure
{
P pred_;
constexpr trim_view_range_adaptor_closure(P pred)
: pred_(pred)
{}
template <rg::viewable_range R>
constexpr auto operator()(R && r) const
{
return trim_view(std::forward<R>(r), pred_);
}
} ;
struct trim_view_range_adaptor
{
template<rg::viewable_range R, typename P>
constexpr auto operator () (R && r, P pred)
{
return trim_view( std::forward<R>(r), pred ) ;
}
template <typename P>
constexpr auto operator () (P pred)
{
return trim_view_range_adaptor_closure(pred);
}   
};
template <rg::viewable_range R, typename P>
constexpr auto operator | (R&& r, trim_view_range_adaptor_closure<P> const & a)
{
return a(std::forward<R>(r)) ;
}
}
namespace views
{
inline static details::trim_view_range_adaptor trim;
}

它工作正常。我写了一些测试来确保它没问题。

template <typename P>
void are_equal(std::vector<int> const & input, std::vector<int> const & output, P&& pred)
{
std::size_t index = 0;
for(auto  i : input | views::trim(std::forward<P>(pred)))
{
assert(i == output[index]);
index++;
}
assert(index == output.size());
}
int main()
{
auto is_odd = [](const int x){return x%2==1;};
are_equal({}, {}, is_odd);
are_equal({1}, {}, is_odd);    
are_equal({1,3,5}, {}, is_odd);
are_equal({2}, {2}, is_odd);
are_equal({2,4}, {2,4}, is_odd);
are_equal({2,3,4}, {2,3,4}, is_odd);
are_equal({1,2,3,4}, {2,3,4}, is_odd);
are_equal({1,1,2,3,4}, {2,3,4}, is_odd);
are_equal({2,3,4,5}, {2,3,4}, is_odd);
are_equal({2,3,4,5,5}, {2,3,4}, is_odd);
are_equal({1,2,3,4,5}, {2,3,4}, is_odd);
are_equal({1,1,2,3,4,5,5}, {2,3,4}, is_odd);
}

问题是，当我在修剪后应用views::reverse视图时，它不再正常工作。

template <typename P>
void are_equal_reverse2(std::vector<int> const & input, std::vector<int> const & output, P&& pred)
{
std::size_t index = 0;
for(auto  i : input | views::trim(std::forward<P>(pred)) | rg::views::reverse)
{
assert(i == output[index]);
index++;
}
assert(index == output.size());
}
int main()
{
auto is_odd = [](const int x){return x%2==1;};
// OK
are_equal_reverse2({}, {}, is_odd);
are_equal_reverse2({1}, {}, is_odd);
are_equal_reverse2({1,3,5}, {}, is_odd);
are_equal_reverse2({2}, {2}, is_odd);
are_equal_reverse2({2,4}, {4,2}, is_odd);
are_equal_reverse2({2,3,4}, {4,3,2}, is_odd);
are_equal_reverse2({1,2,3,4}, {4,3,2}, is_odd);
are_equal_reverse2({1,1,2,3,4}, {4,3,2}, is_odd);
// fail
are_equal_reverse2({2,3,4,5}, {4,3,2}, is_odd);
are_equal_reverse2({2,3,4,5,5}, {4,3,2}, is_odd);
are_equal_reverse2({1,2,3,4,5}, {4,3,2}, is_odd);
are_equal_reverse2({1,1,2,3,4,5,5}, {4,3,2}, is_odd);
}

范围 {2,3,4,5}变为 {2,3,4}。反向应用后，它应变为 {4,3,2}。但是，结果实际上是 {5,4,3,2}。

我希望views::reversestd::make_reverse_iterator()应用于修剪视图的开始和结束迭代器。这应执行以下转换：

trim_view        reverse_view (expected)      reverse_view (actual)
--------------------------------------------------------------------
2 3 4 5 _         _ 2 3 4 5                   _ 2 3 4 5
^     ^           ^     ^                     ^       ^
|     |      =>   |     |                     |       |
|     end_        rend  |                     rend    |
iter_                   rbegin                        rbegin

我不确定我在这里错过了什么。任何帮助，不胜感激。

以下是工作示例的链接：https://wandbox.org/permlink/4iFNsqiz9Y9Bfm64

Answer 1

首先，让我们从这个开始：

constexpr auto begin()
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() const requires rg::range<const R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() requires rg::random_access_range<R> && rg::sized_range<R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto begin() const requires rg::random_access_range<const R> && rg::sized_range<const R>
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto end()
{ return end_ ; }
constexpr auto end() const requires rg::range<const R>
{ return end_ ; }
constexpr auto end() requires rg::random_access_range<R> && rg::sized_range<R>
{ return end_ ; }
constexpr auto end() const requires rg::random_access_range<const R> && rg::sized_range<const R>
{ return end_ ; }

所有这些重载都做完全相同的事情，所以让我们减少到两个：

constexpr auto begin() const
{
while(iter_ != std::end(base_) && pred_(*iter_)) {iter_ = std::next(iter_);} 
while(end_ != iter_ && pred_(*std::prev(end_))) {end_ = std::prev(end_);}
return iter_;
}
constexpr auto end() const
{ return end_ ; }

好吧。这是怎么回事？begin()将调整要修剪的iter_和end_，end()只是返回end_。

这很好，如果你这样做：

auto trimmed = some_range | trim(some_pred);
auto b = trimmed.begin();
auto e = trimmed.end();

但是，如果您这样做会发生什么：

auto e = trimmed.end();
auto b = trimmed.begin();

end在这种情况下将是some_range.end()，它不会是这个范围的正确结束迭代器！您需要确保begin()和end()之间没有任何排序依赖关系 - 它们始终必须返回正确的值。

<小时 />

此外，trim(p)可以全部减少为：

template <rg::viewable_range R, typename P>
constexpr auto operator ()(R && r, P pred)
{
auto negated = std::not_fn(pred);
auto f = rg::find_if(r, negated);
auto l = rg::find_if(r | std::views::reverse, negated).base();
return rg::subrange(f, l);
}