c++ - std::map 的查找失败，但手动扫描 map 中的元素

Question

我正在使用 std::map 和一个列表来跟踪元素和相关分数的窗口。当窗口已满时，我想从窗口队列中弹出一个元素并将其从 map 中删除。因为可能存在重复项， map 会跟踪窗口中每个元素出现的次数。我还使用有序映射，以便我可以在给定窗口中不断获得最小值。

我的问题是 find() 返回了 end()，但它不应该返回。当我遍历 map 时，我发现该元素存在。我不想牺牲使用 map 的对数复杂度。

tl;dr:std::map 表示一个元素不在 map 中。手动扫描说是。

[编辑:Bryan Chen 的建议修复了 map 。谢谢!]

#include <cstdint>
#include <cstdio>
#include <cinttypes>
#include <map>
#include <list>
#include <vector>

#include "util.h"
#include "kmerutil.h"

namespace kpg {

struct elscore_t {
    uint64_t el_, score_;
    INLINE elscore_t(uint64_t el, uint64_t score): el_(el), score_(score) {
        LOG_ASSERT(el == el_);
        LOG_ASSERT(score == score_);
    }
    INLINE elscore_t(): el_(0), score_(0) {}
    inline bool operator <(const elscore_t &other) const {
        return score_ < other.score_ || el_ < other.el_; // Lexicographic is tie-breaker.
    }
    inline bool operator ==(const elscore_t &other) const {
        return score_ == other.score_ && el_ == other.el_; // Lexicographic is tie-breaker.
    }
    std::string to_string() const {
        return std::to_string(el_) + "," + std::to_string(score_);
    }
};

struct esq_t: public std::list<elscore_t> {
};

typedef std::map<elscore_t, unsigned> esmap_t;

class qmap_t {
    // I could make this more efficient by using pointers instead of
    // elscore_t structs.
    // *maybe* TODO
    // Could also easily templatify this module for other windowing tasks.
    esq_t list_;
#if !NDEBUG
public:
    esmap_t map_;
private:
#else
    esmap_t map_;
#endif
    const size_t wsz_;  // window size to keep
    public:
    void add(const elscore_t &el) {
        auto it(map_.upper_bound(el));
        if(it->first == el) ++it->second;
        else map_.emplace(el, 1);
    }
    void del(const elscore_t &el) {
        auto f(map_.find(el));
        if(f == map_.end()) {
            LOG_DEBUG("map failed :(\n");
            for(f = map_.begin(); f != map_.end(); ++f)
                if(f->first == el)
                    break;
        }
        LOG_ASSERT(f != map_.end());
        if(--f->second <= 0)
            map_.erase(f);
    }
    uint64_t next_value(const uint64_t el, const uint64_t score) {
        list_.emplace_back(el, score);
        LOG_ASSERT(list_.back().el_ == el);
        LOG_ASSERT(list_.back().score_ == score);
        add(list_.back());
        if(list_.size() > wsz_) {
            //fprintf(stderr, "list size: %zu. wsz: %zu\n", list_.size(), wsz_);
            //map_.del(list_.front());
            del(list_.front());
            list_.pop_front();
        }
        LOG_ASSERT(list_.size() <= wsz_);
        return list_.size() == wsz_ ? map_.begin()->first.el_: BF;
        // Signal a window that is not filled by 0xFFFFFFFFFFFFFFFF
    }
    qmap_t(size_t wsz): wsz_(wsz) {
    }
    void reset() {
        list_.clear();
        map_.clear();
    }
};

}

Answer 1

这不是有效的严格弱排序:

return score_ < other.score_ || el_ < other.el_;

你有 elscore_t(0, 1) < elscore_t(1, 0)和 elscore_t(1, 0) < elscore_t(0, 1) .