问题描述
假设我有以下代码:
vector<int> list;
for(auto& elem:list) {
int i = elem;
}
我可以在不维护单独迭代器的情况下找到 elem
在向量中的位置吗?
Can I find the position of elem
in the vector without maintaining a separate iterator?
推荐答案
是的,只是需要一些按摩;)
Yes you can, it just take some massaging ;)
诀窍是使用组合:您无需直接遍历容器,而是在整个过程中使用索引压缩"它.
The trick is to use composition: instead of iterating over the container directly, you "zip" it with an index along the way.
专业的邮政编码:
template <typename T>
struct iterator_extractor { typedef typename T::iterator type; };
template <typename T>
struct iterator_extractor<T const> { typedef typename T::const_iterator type; };
template <typename T>
class Indexer {
public:
class iterator {
typedef typename iterator_extractor<T>::type inner_iterator;
typedef typename std::iterator_traits<inner_iterator>::reference inner_reference;
public:
typedef std::pair<size_t, inner_reference> reference;
iterator(inner_iterator it): _pos(0), _it(it) {}
reference operator*() const { return reference(_pos, *_it); }
iterator& operator++() { ++_pos; ++_it; return *this; }
iterator operator++(int) { iterator tmp(*this); ++*this; return tmp; }
bool operator==(iterator const& it) const { return _it == it._it; }
bool operator!=(iterator const& it) const { return !(*this == it); }
private:
size_t _pos;
inner_iterator _it;
};
Indexer(T& t): _container(t) {}
iterator begin() const { return iterator(_container.begin()); }
iterator end() const { return iterator(_container.end()); }
private:
T& _container;
}; // class Indexer
template <typename T>
Indexer<T> index(T& t) { return Indexer<T>(t); }
并使用它:
#include <iostream>
#include <iterator>
#include <limits>
#include <vector>
// Zipper code here
int main() {
std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9};
for (auto p: index(v)) {
std::cout << p.first << ": " << p.second << "\n";
}
}
您可以在 ideone 上看到它,尽管它缺少for-range循环支持,因此不太美观.
You can see it at ideone, though it lacks the for-range loop support so it's less pretty.
只记得我应该更频繁地检查Boost.Range.不幸的是,没有 zip
范围,但是我确实找到了一个perl:.但是,它需要访问迭代器才能提取索引.丢人:x
Just remembered that I should check Boost.Range more often. Unfortunately no zip
range, but I did found a perl: boost::adaptors::indexed
. However it requires access to the iterator to pull of the index. Shame :x
否则,使用 counting_range
和通用的 zip
我相信可以做一些有趣的事情...
Otherwise with the counting_range
and a generic zip
I am sure it could be possible to do something interesting...
在理想世界中,我会想象:
In the ideal world I would imagine:
int main() {
std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9};
for (auto tuple: zip(iota(0), v)) {
std::cout << tuple.at<0>() << ": " << tuple.at<1>() << "\n";
}
}
使用 zip
自动创建一个视图,作为引用元组的范围,而 iota(0)
只需创建一个从 0
,并且仅计入无穷大(或者,它的最大值……).
With zip
automatically creating a view as a range of tuples of references and iota(0)
simply creating a "false" range that starts from 0
and just counts toward infinity (or well, the maximum of its type...).
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