std::ranges::next
From cppreference.com
Defined in header <iterator>
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Call signature |
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template
<
std::input_or_output_iterator I >
constexpr I next( I i ) ; |
(1) | (since C++20) |
template
<
std::input_or_output_iterator I >
constexpr I next( I i, std::iter_difference_t <I> n ) ; |
(2) | (since C++20) |
template
<
std::input_or_output_iterator I, std::sentinel_for
<I> S >
constexpr I next( I i, S bound ) ; |
(3) | (since C++20) |
template
<
std::input_or_output_iterator I, std::sentinel_for
<I> S >
constexpr I next( I i, std::iter_difference_t <I> n, S bound ) ; |
(4) | (since C++20) |
Return the nth successor of iterator i.
The function-like entities described on this page are algorithm function objects (informally known as niebloids), that is:
- Explicit template argument lists cannot be specified when calling any of them.
- None of them are visible to argument-dependent lookup.
- When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup
Parameters
i | - | an iterator |
n | - | number of elements to advance |
bound | - | sentinel denoting the end of the range i points to |
Return value
1) The successor of iterator i.
2) The nth successor of iterator i.
3) The first iterator equivalent to bound.
4) The n
th successor of iterator i, or the first iterator equivalent to bound
Complexity
1) Constant.
3) Constant if
I
and S
models both
std::random_access_iterator<I>
and
std::sized_sentinel_for<S, I>
, or if I
and S
models
std::assignable_from
<I&, S>
Possible implementation
struct next_fn { template<std::input_or_output_iterator I> constexpr I operator()(I i) const { ++i; return i; } template<std::input_or_output_iterator I> constexpr I operator()(I i, std::iter_difference_t<I> n) const { ranges::advance(i, n); return i; } template<std::input_or_output_iterator I, std::sentinel_for<I> S> constexpr I operator()(I i, S bound) const { ranges::advance(i, bound); return i; } template<std::input_or_output_iterator I, std::sentinel_for<I> S> constexpr I operator()(I i, std::iter_difference_t<I> n, S bound) const { ranges::advance(i, n, bound); return i; } }; inline constexpr auto next = next_fn(); |
Notes
Although the expression ++x.begin() often compiles, it is not guaranteed to do so:
x.begin()
is an rvalue expression, and there is no 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, ++x.begin() does not compile, while
ranges::next(x.begin())
Example
Run this code
#include <cassert> #include <iterator> int main() { auto v = {3, 1, 4}; { auto n = std::ranges::next(v.begin()); assert(*n == 1); } { auto n = std::ranges::next(v.begin(), 2); assert(*n == 4); } { auto n = std::ranges::next(v.begin(), v.end()); assert(n == v.end()); } { auto n = std::ranges::next(v.begin(), 42, v.end()); assert(n == v.end()); } }
See also
(C++20)
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decrement an iterator by a given distance or to a bound (algorithm function object) |
(C++20)
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advances an iterator by given distance or to a given bound (algorithm function object) |
(C++11)
|
increment an iterator (function template) |