我编写了一个C ++ 20范围视图（不是range-v3视图）trim，该视图给定了范围和一元谓词，返回了一个新的范围，而没有满足谓词的前后元素。（range-v3库具有这样的视图，但是C ++ 20中缺少它。）
这是实现（这可能不是最佳方法，但是ranges库上没有很多文档，因此在资源有限的情况下，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。我写了一些测试以确保一切正常。
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views::reverse在修剪后应用视图时，它将不再正常工作。
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}。启用反向后，它将变为{4,3,2}。但是，结果实际上是{5,4,3,2}。
我希望这views::reverse适用std::make_reverse_iterator()于修剪视图的开始和结束迭代器。那应该执行以下转换：
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
我不确定我在这里缺少什么。任何帮助表示赞赏。
这是工作示例的链接：https : //wandbox.org/permlink/4iFNsqiz9Y9Bfm64
首先，让我们开始：
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_ ; }
所有这些重载都执行完全相同的操作，因此我们可以减少到两个：
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 这里发生了什么？begin()将同时调整iter_和end_进行修剪，然后end()返回end_。
很好，如果您这样做：
auto trimmed = some_range | trim(some_pred);
auto b = trimmed.begin();
auto e = trimmed.end();
但是，如果执行此操作会发生什么：
auto e = trimmed.end();
auto b = trimmed.begin();
end在这种情况下将是some_range.end()，它将不是该范围的正确的最终迭代器！你需要确保begin()和end()没有他们之间的顺序依赖-他们总是要返回正确的值。
同样，trim(p)可以将其整体减少为：
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);
}
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