我有一个实现< code>begin和< code>end的自定义容器。如何将这个容器通过管道连接到ranges-v3视图？
std：：vector是可管道的，所以我尝试用同样的方法来管道化我的自定义类，但在我的容器中找不到管道操作符。
我查阅了文档，但是除了用Range接口重新实现一个包装类之外，我找不到任何其他方法。我有多个这样的类，我相信这可能是一个相当常见的情况，所以我宁愿使用库提供的一些函数(或类库)，但我无法从文档中找到它。
下面是一个最小示例：
#include <iostream>
#include <iterator>
#include <range/v3/all.hpp>
struct Test {
struct iterator;
struct sentinel {};
int counter;
Test() = default;
iterator begin();
sentinel end() const { return {}; }
iterator begin() const;
};
struct Test::iterator {
using value_type = int;
using reference = int&;
using pointer = int*;
using iterator_category = std::input_iterator_tag;
using difference_type = void;
Test* test;
iterator& operator++() {
test->counter++;
return *this;
}
iterator operator++(int) {
auto it = *this;
++*this;
return it;
}
int operator*() { return test->counter; }
int operator*() const { return test->counter; }
bool operator!=(const iterator& rhs) const {
return rhs.test != test;
}
bool operator!=(sentinel) const {
return true;
}
};
Test::iterator Test::begin() { return iterator {this}; }
Test::iterator Test::begin() const { return iterator {const_cast<Test*>(this)}; }
int main() {
auto container = Test();
static_assert(ranges::range<Test>, "It is not a range");
static_assert(ranges::viewable_range<Test>, "It is not a viewable range");
auto rng = container | ranges::views::take(10);
for (auto n : rng) { std::cerr << n << std::endl;}
return 0;
}
这是我使用此代码时遇到的错误：
~/tmp/range$ g++ main.cpp -Irange-v3/include -o main 2>&1 | grep error
main.cpp:46:19: error: static assertion failed: It is not a range
main.cpp:47:19: error: static assertion failed: It is not a viewable range
range-v3/include/range/v3/functional/pipeable.hpp:63:53: error: no matching function for call to ‘ranges::pipeable_access::impl<ranges::views::view<ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::take_fn, int>]::_> >::pipe(Test&, ranges::views::view<ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::take_fn, int>]::_>&)’
main.cpp:48:10: error: ‘void rng’ has incomplete type
main.cpp:49:19: error: unable to deduce ‘auto&&’ from ‘rng’
这是我的解决方案：
template <typename T>
class ContainerView : public ranges::view_facade<ContainerView<T>> {
friend ranges::range_access;
using iterator_type = decltype(std::declval<T>().begin());
using value_type = decltype(*std::declval<iterator_type>());
T* container_;
iterator_type it_;
value_type read() const { return *it_; }
bool equal(ranges::default_sentinel_t) const {
return !(it_ != container_->end());
}
void next() { ++it_; }
public:
ContainerView() = default;
ContainerView(T& container)
: container_(&container), it_(container_->begin()) {}
using ranges::view_facade<ContainerView<T>>::begin;
using ranges::view_facade<ContainerView<T>>::end;
};
int main() {
auto container = Test();
auto view = ContainerView(container);
static_assert(ranges::range<decltype(view)>, "It is not a range");
static_assert(ranges::viewable_range<decltype(view)>,
"It is not a viewable range");
auto rng = view | ranges::views::take(10);
for (auto n : rng) {
std::cerr << n << std::endl;
}
return 0;
}
我不确定我是否重新实现了库中已经实现的东西。所以，我暂时不会将此标记为答案。如果有现有的方法，请评论或发布新的答案。
对sentinel_for进行建模，迭代器和哨兵必须在两个方向上使用 == 和 ！= 进行比较。（输入迭代器在新世界中不必相互比较。您只提供了 ！= 并且仅在一个方向上提供。difference_type对于输入迭代ptrdiff_t。此外：
引用应该是运算符 * 的返回类型;它不必是引用类型。viewable_range