如何推导出std :: advance迭代器类型？

Question

Please take a look at the std::advance function. According to cppreference the complexity is:

Linear. However, if InputIt additionally meets the requirements of RandomAccessIterator, complexity is constant.

So if I pass an iterator and an integer, how does the program deduce what iterator it is and how it is implemented? So are there 2 overloads for the function or what?

Answer 1

how program deduce what iterator it and how it is implemented?

It uses std::iterator_traits<T>::iterator_category to determine iterator category and then does tag dispatching .

std::iterator_traits is specialised for raw pointers. Raw pointers are random-access iterators.

Eg:

template<class I>
void doadvance(I& i, size_t n, std::random_access_iterator_tag) {
    i += n;
}

template<class I, class Tag>
void doadvance(I& i, size_t n, Tag) {
    while(n--)
        ++i;
}

template<class I>
void advance(I& i, size_t n) {
    using Tag = typename std::iterator_traits<I>::iterator_category;
    doadvance(i, n, Tag{});
}

In C++17 instead of tag dispatching it can use if constexpr , but that would make the standard library not backward compatible. This is why it has to keep using tag dispatching .

IMO, std::next has a better interface than std::advance because it returns the iterator, so that the invocation does not have to occupy its own line of code.

Answer 2

how program deduce what iterator it is

std::iterator_traits is used, ie std::iterator_traits<It>::iterator_category will be checked, and if it's std::random_access_iterator_tag , means it's a RandomAccessIterator.

BTW: std::iterator_traits is specialized for raw pointers; they're always RandomAccessIterator.

So are there 2 overloads for the function or what?

There're no more overloads of std::advance , the implementation usually implements several helper function overloads, which are called according to the iterator_category in std::advance .