C++14生成自定义迭代器,该迭代器将按2进行运算并返回修改后的数据

C++14生成自定义迭代器,该迭代器将按2进行运算并返回修改后的数据,c++,iterator,c++14,C++,Iterator,C++14,我有一个类范围,它包含一个类似或的间隔 然后我有一个范围列表,我想要一个代表多个范围的向量。出于某种目的,我不需要向量 但我想做一个迭代器,它将遍历范围并重新运行范围。可以定义这样工作的自定义迭代器吗 class Range { long long lo; long long hi; } class RangeList { vector<long long> ranges; } 所以迭代器将遍历并返回 第一次

我有一个类范围,它包含一个类似或的间隔 然后我有一个范围列表,我想要一个代表多个范围的向量。出于某种目的,我不需要向量

但我想做一个迭代器,它将遍历范围并重新运行范围。可以定义这样工作的自定义迭代器吗

   class Range {
        long long lo;
        long long hi;
    }

    class RangeList {
    vector<long long> ranges;

    }
所以迭代器将遍历并返回

第一次迭代

Range <1,50>
二次畸形

Range <200,700>
第三次畸形

Range <900,100>

谢谢你的建议

我想我应该更具体一点,因为我不想从连续的值列表中推断成对的值,如果有奇数个值会发生什么

#include <vector>
#include <iostream>

template<class Integer>
struct InclusiveRangeIter
{
    using iterator_category = std::forward_iterator_tag;
    using value_type = Integer;
    using reference = value_type&;
    using pointer = value_type*;
    using difference_type = Integer;

    constexpr InclusiveRangeIter(Integer current)
    : value_(current)
    {}

    constexpr bool operator==(InclusiveRangeIter const& other) const { return value_ == other.value_; }
    constexpr bool operator!=(InclusiveRangeIter const& other) const { return value_ != other.value_; }
    value_type operator*() const { return value_; }
    auto operator++() -> InclusiveRangeIter& { ++value_; return *this; }
    auto operator++(int) -> InclusiveRangeIter { auto copy = *this; ++value_; return copy; }


    Integer value_;
};

struct InclusiveRange 
{
    long long lo;
    long long hi;

    auto begin() const { return InclusiveRangeIter(lo); }
    auto end() const { return InclusiveRangeIter(hi + 1); }
};

int main()
{
    auto ranges = std::vector<InclusiveRange>
    {
        {1, 50}, {200, 700}, {900, 1000}
    };

    for (auto&& ir : ranges)
    {
        auto sep = false;
        for (auto&& v : ir)
        {
            if (sep) std::cout << ", ";
            std::cout << v;
            sep = true;
        }
        std::cout << '\n';
    }
}

是的,这是可能的-如何:boost::tranform_迭代器?这是可能的,但可能适得其反。你为什么不想要一个向量?我很想使用Ranges=std::vector;,和sidestep RangeListOk谢谢你们的帮助,我会像你们在vector中那样做,我不允许使用boost,但你们是对的:谢谢你们
#include <vector>
#include <iostream>

template<class Integer>
struct InclusiveRangeIter
{
    using iterator_category = std::forward_iterator_tag;
    using value_type = Integer;
    using reference = value_type&;
    using pointer = value_type*;
    using difference_type = Integer;

    constexpr InclusiveRangeIter(Integer current)
    : value_(current)
    {}

    constexpr bool operator==(InclusiveRangeIter const& other) const { return value_ == other.value_; }
    constexpr bool operator!=(InclusiveRangeIter const& other) const { return value_ != other.value_; }
    value_type operator*() const { return value_; }
    auto operator++() -> InclusiveRangeIter& { ++value_; return *this; }
    auto operator++(int) -> InclusiveRangeIter { auto copy = *this; ++value_; return copy; }


    Integer value_;
};

struct InclusiveRange 
{
    long long lo;
    long long hi;

    auto begin() const { return InclusiveRangeIter(lo); }
    auto end() const { return InclusiveRangeIter(hi + 1); }
};

int main()
{
    auto ranges = std::vector<InclusiveRange>
    {
        {1, 50}, {200, 700}, {900, 1000}
    };

    for (auto&& ir : ranges)
    {
        auto sep = false;
        for (auto&& v : ir)
        {
            if (sep) std::cout << ", ";
            std::cout << v;
            sep = true;
        }
        std::cout << '\n';
    }
}