使用基于范围的for循环时需要迭代器

Question

目前,我只能使用以下方法进行基于范围的循环:

for (auto& value : values)

但有时我需要一个值的迭代器,而不是一个引用(无论出于何种原因).是否有任何方法无需通过整个矢量比较值？

Answer 1

使用旧for循环:

for (auto it = values.begin(); it != values.end();  ++it )
{
       auto & value = *it;
       //...
}

有了这个,你value就像迭代器一样it.使用你想要使用的任何东西.

编辑:

虽然我不推荐这个,但如果你想使用基于范围的for循环(是的,无论出于何种原因:D),那么你可以这样做:

 auto it = std::begin(values); //std::begin is a free function in C++11
 for (auto& value : values)
 {
     //Use value or it - whatever you need!
     //...
     ++it; //at the end OR make sure you do this in each iteration
 }

这种方法避免了搜索value,因为value并且it始终保持同步.

Answer 2

这是一个代理包装类,允许您通过将其别名化为您自己的变量来公开隐藏的迭代器.

#include <memory>
#include <iterator>

/*  Only provides the bare minimum to support range-based for loops.
    Since the internal iterator of a range-based for is inaccessible,
    there is no point in more functionality here. */
template< typename iter >
struct range_iterator_reference_wrapper
    : std::reference_wrapper< iter > {
    iter &operator++() { return ++ this->get(); }
    decltype( * std::declval< iter >() ) operator*() { return * this->get(); }
    range_iterator_reference_wrapper( iter &in )
        : std::reference_wrapper< iter >( in ) {}
    friend bool operator!= ( range_iterator_reference_wrapper const &l,
                             range_iterator_reference_wrapper const &r )
        { return l.get() != r.get(); }
};

namespace unpolluted {
    /*  Cannot call unqualified free functions begin() and end() from 
        within a class with members begin() and end() without this hack. */
    template< typename u >
    auto b( u &c ) -> decltype( begin( c ) ) { return begin( c ); }
    template< typename u >
    auto e( u &c ) -> decltype( end( c ) ) { return end( c ); }
}

template< typename iter >
struct range_proxy {
    range_proxy( iter &in_first, iter in_last )
        : first( in_first ), last( in_last ) {}

    template< typename T >
    range_proxy( iter &out_first, T &in_container )
        : first( out_first ),
        last( unpolluted::e( in_container ) ) {
        out_first = unpolluted::b( in_container );
    }

    range_iterator_reference_wrapper< iter > begin() const
        { return first; }
    range_iterator_reference_wrapper< iter > end()
        { return last; }

    iter &first;
    iter last;
};

template< typename iter >
range_proxy< iter > visible_range( iter &in_first, iter in_last )
    { return range_proxy< iter >( in_first, in_last ); }

template< typename iter, typename container >
range_proxy< iter > visible_range( iter &first, container &in_container )
    { return range_proxy< iter >( first, in_container ); }

用法:

#include <vector>
#include <iostream>
std::vector< int > values{ 1, 3, 9 };

int main() {
    // Either provide one iterator to see it through the whole container...
    std::vector< int >::iterator i;
    for ( auto &value : visible_range( i, values ) )
        std::cout << "# " << i - values.begin() << " = " << ++ value << '\n';

    // ... or two iterators to see the first incremented up to the second.
    auto j = values.begin(), end = values.end();
    for ( auto &value : visible_range( j, end ) )
        std::cout << "# " << j - values.begin() << " = " << ++ value << '\n';
}

Answer 3

我尝试了这个并找到了解决方案.

用法:

for(auto i : ForIterator(some_list)) {
    // i is the iterator, which was returned by some_list.begin()
    // might be useful for whatever reason
}

实施并不困难:

template <typename T> struct Iterator {
    T& list;
    typedef decltype(list.begin()) I;

    struct InnerIterator {
        I i;
        InnerIterator(I i) : i(i) {}
        I operator * () { return i; }
        I operator ++ () { return ++i; }
        bool operator != (const InnerIterator& o) { return i != o.i; }
    };

    Iterator(T& list) : list(list) {}
    InnerIterator begin() { return InnerIterator(list.begin()); }
    InnerIterator end() { return InnerIterator(list.end()); }
};
template <typename T> Iterator<T> ForIterator(T& list) {
    return Iterator<T>(list);
}