我想知道 C++20 范围是否有一些很好的方法让我迭代排序容器的相等范围(或者在一般情况下任何排序范围)。
我有这份“手册”solution这行得通,但从某种意义上说它并不好,因为它不是真正可组合的(我没有得到一些相等范围 View 的 View ,我只是提供要调用的函数)。
#include <vector>
#include <ranges>
#include <iostream>
#include <fmt/ranges.h>
template<typename Cont, typename Fn>
void for_each_equal_range(const Cont& cont, Fn fn){
auto current_begin = cont.begin();
while(true){
if (current_begin==cont.end()){
return;
}
auto [eq_begin, eq_end] = std::equal_range(current_begin, cont.end(), *current_begin);
fn(eq_begin, eq_end);
current_begin = eq_end;
}
}
int main() {
std::vector vals {1,2,2,3,3,3,47};
for_each_equal_range(vals, [](const auto b, const auto e){
std::cout << "size: " << std::distance(b,e) << std::endl;
std::ranges::subrange elems(b, e);
std::cout << fmt::format("{}",elems) << std::endl;
});
}
我希望我有这样的东西:
vals | equal_range_split | std::ranges::for_each(...);
如果对此处的范围含义感到困惑:
- 相等范围是很好的旧 STL equal_range 含义
- ranges 是 C++20 范围库。
我也知道 C++20 有 std::ranges::equal_range
算法,但它似乎对我的用例没有太大帮助。
最佳答案
不完全是你要找的东西,但如果你有协程 generator
模板,比如
using namespace std::ranges;
template<typename Range, typename Compare = std::less>
generator<subrange<iterator_t<Range>>> equal_ranges(Range&& range, Compare compare = {}) {
for (auto current_begin = cont.begin(); current_begin != cont.end();) {
auto [eq_begin, eq_end] = std::equal_range(current_begin, cont.end(), *current_begin, compare);
co_yield { eq_begin, eq_end };
current_begin = eq_end;
}
}
template<typename Compare>
struct equal_ranges_holder {
Compare compare;
};
template<typename Compare = std::less>
equal_ranges_holder<Compare> equal_ranges(Compare compare = {}) { return { compare }; }
template<typename Range, typename Compare>
auto operator|(Range&& range, equal_ranges_holder<Compare> holder) {
return equal_ranges(range, holder.compare);
}
或者,进行单次传递 equal_range_view
也相当简单。
using namespace std::ranges;
template<borrowed_range Range, std::strict_weak_order Compare = std::less>
class equal_range_view : public view_interface<equal_range_view> {
using base_iterator = iterator_t<Range>;
using base_sentinel = sentinel_t<Range>;
Range range;
Compare compare;
public:
class sentinel {};
class iterator {
base_iterator base;
base_sentinel end;
Compare compare;
public:
using value_type = subrange<base_iterator>;
using reference = value_type;
using pointer = value_type *;
using difference_type = std::ptrdiff_t;
using iterator_category = std::input_iterator_tag;
equal_range_iterator(base_iterator base, base_sentinel end, Compare compare) : base(base), end(end), compare(compare) {}
reference operator*() {
auto [first, last] = std::equal_range(base, end, compare);
return { first, last };
}
iterator& operator++() {
auto [_, last] = std::equal_range(base, end, compare);
base = last;
return *this;
}
bool operator==(sentinel) const {
return base == end;
}
bool operator!=(sentinel) const {
return base != end;
}
};
explicit equal_range_view(Range range, Compare compare = {}) : range(range), compare(compare) {}
iterator begin() { return { range.begin(), range.end(), compare }; }
sentinel end() { return {}; }
};
关于c++ - 使用 C++20 遍历相等的范围,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/68448698/