c++ - boost multi_index 反向迭代器删除麻烦

Question

我有以下(简化的)代码:

#include <boost/multi_index_container.hpp>
#include <boost/multi_index/ordered_index.hpp>
namespace bmi = boost::multi_index;

#include <string>
#include <iostream>
#include <cassert>

using Container = boost::multi_index_container<
    std::string,
    bmi::indexed_by< bmi::ordered_non_unique< bmi::identity<std::string> > >
>;

/// Get the base of a non-reverse iterator. It's the iterator itself.
inline
Container::iterator const&
iter_base(Container::iterator const& it)
{
    return it;
}

/** Get a non-reverse iterator that points at the same element as the given reverse_iterator.
 *
 * @param rit reverse_iterator
 * @return a (non-reverse) iterator that points to the same element.
 * @pre @p rit is dereferenceable (not equal to @c rend() of whatever container @p rit came from)
 */
inline
Container::iterator
iter_base(Container::reverse_iterator const& rit)
{
    auto bit = rit.base();
    // if 'rit' is a reverse iterator: &*(rit.base() - 1) == &*rit
    return --bit;
}

template <typename IT>
void evict(Container& c, IT rb, IT fin)
{
    std::vector<std::string> result;
    for (; rb != fin; ) {
        if (rb->size() == 3) {
            auto victim = rb;
            ++rb;
            std::cout << "victim->" << *victim << ", next->" << (rb==fin ? std::string{"THE END"} : *rb) << "\n";
            auto next = c.erase(iter_base(victim));
            std::cout << "size=" << c.size() << "\n";
            for (auto const& s : c) {
                std::cout << "remain: " << s << "\n"; // bar - baz - foo
            }

            rb = IT(next);
            (void)next;
        }
        else {
            result.push_back(*rb);
        }
    }
}

int main(int argc, char**)
{
    bool forward = (argc == 1);

    Container c;
    c.insert("foo"); // will be last
    c.insert("bar");
    c.insert("baz");

    if (forward) {
        auto b = c.lower_bound("baz");

        std::cout << ">> " << *b << "\n"; // prints baz

        auto rb = (b);
        std::cout << "<< " << *rb            << "\n"; // prints baz
        std::cout << "<< " << *iter_base(rb) << "\n"; // prints baz

        evict(c, rb, c.end());
    }
    else {
        auto b = c.upper_bound("baz");

        std::cout << ">> " << *b << "\n"; // prints foo

        auto rb = Container::reverse_iterator(b);
        std::cout << "<< " << *rb            << "\n"; // prints baz
        std::cout << "<< " << *iter_base(rb) << "\n"; // prints baz

        evict(c, rb, c.rend());
    }
}

真正的代码不仅仅是删除，但这足以说明行为。

已编辑以表明在循环中不会发生任何删除。根据使用的迭代器类型，项目应该以正向或反向顺序添加到 result。

当不带参数运行时，forward==true 并且输出符合预期:

>> baz
<< baz
<< baz
victim->baz, next->foo
size=2
remain: bar
remain: foo
victim->foo, next->THE END
size=1
remain: bar

当使用参数运行时，forward==false 输出为:

>> foo
<< baz
<< baz
victim->baz, next->bar
size=2
remain: bar
remain: foo
segmentation fault (core dumped)

(不符合预期)

使用地址清理器编译会在第 42 行(++rb 行)显示堆释放后使用。

似乎调用 erase(victim) 以某种方式使 rb 无效，即使 erase 不应该使任何其他迭代器无效。

知道我做错了什么吗？

Answer 1

第二个答案，OP 的额外要求是根据迭代器的性质以正序或逆序进行遍历。稍微小心一点，可以这样做:<罢工>

<罢工>

template <typename IT>
void evict(Container& c, IT rb, IT fin)
{
    std::vector<std::string> result;
    if(rb != fin) for(;;) {
        IT next = rb;
        ++next;
        bool finished  = (next == fin);
        if (rb->size() == 3) {
            c.erase(iter_base(rb));
            std::cout << "size=" << c.size() << "\n";
            for (auto const& s : c) {
                std::cout << "remain: " << s << "\n"; // bar - baz - foo
            }
        }
        else {
            result.push_back(*rb);
        }
        if(finished) break;
        rb = next;
    }
}

<罢工> 糟糕的是，受灾代码仍在运行到 UB。请试试这个:

template <typename IT>
void evict(Container& c, IT rb, IT fin)
{
    std::vector<std::string> result;
    if(rb != fin) for(;;) {
        bool finished  = (std::next(rb) == fin);
        if (rb->size() == 3) {
            rb = IT{c.erase(iter_base(rb))};
            std::cout << "size=" << c.size() << "\n";
            for (auto const& s : c) {
                std::cout << "remain: " << s << "\n"; // bar - baz - foo
            }

        }
        else {
            result.push_back(*rb);
        }
        if(finished) break;
    }
}