std::next(3)

💅 std::next(3) — C++ Standard Libary, bestie 💅

okay so welcome to std::next, the iterator-advancing icon of the C++ Standard Library. This is the std::next(3) manual page, and we're gonna get you comfy with it. no cap. ✨

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🌸 NAME 🌸

std::next - std::next

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✨ Synopsis ✨

Defined in header <iterator>.

The page lays out the synopsis like this, version labels and all — I'm keeping it exactly as it appears:

Defined in header <iterator>
template< class InputIt >
(since C++11)
InputIt next( InputIt it, typename
(until C++17)
std::iterator_traits<InputIt>::difference_type n = 1 );
template< class InputIt >

constexpr
(since C++17)

InputIt next( InputIt it, typename
std::iterator_traits<InputIt>::difference_type n = 1 );

Return the n^th successor (or -n^th predecessor if n is negative) of iterator it.

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🍓 Parameters 🍓

• it — an iterator
• n — number of elements to advance

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🎀 Type requirements 🎀

• InputIt must meet the requirements of LegacyInputIterator.

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🦋 Return value 🦋

An iterator of type InputIt that holds the n^th successor (or -n^th predecessor if n is negative) of iterator it.

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⏱️ Complexity ⏱️

Linear.

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

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🛠️ Possible implementation 🛠️

template<class InputIt>
constexpr // since C++17
InputIt next(InputIt it, typename
std::iterator_traits<InputIt>::difference_type n = 1)
{
    std::advance(it, n);
    return it;
}

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📝 Notes 📝

Although the expression ++c.begin() often compiles, it is not guaranteed to do so: c.begin() is an rvalue expression, and there is no LegacyInputIterator requirement that specifies that increment of an rvalue is guaranteed to work. In particular, when iterators are implemented as pointers or its operator++ is lvalue-ref-qualified, ++c.begin() does not compile, while std::next(c.begin()) does.

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🌈 Example 🌈

The page says // Run this code right before the example:

#include <iostream>
#include <iterator>
#include <vector>

int main()
{
    std::vector<int> v{4, 5, 6};

    auto it = v.begin();
    auto nx = std::next(it, 2);
    std::cout << *it << ' ' << *nx << '\n';

    it = v.end();
    nx = std::next(it, -2);
    std::cout << ' ' << *nx << '\n';
}

🖥️ Output:

4 6
 5

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😭 Defect reports 😭

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

• DR LWG 2353 — applied to C++11 — behavior as published: next required LegacyForwardIterator; correct behavior: LegacyInputIterator allowed.

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👀 See also 👀

• prev (C++11) (function template) — decrement an iterator
• advance (function template) — advances an iterator by given distance
• distance (function template) — returns the distance between two iterators
• ranges::next (C++20) (niebloid) — increment an iterator by a given distance or to a bound

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okay that's the whole page, bestie. this one came from http://cppreference.com, dated 2024.06.10, and it's the std::next(3) page. go advance some iterators responsibly 💖✨