std::ranges::subrange<I,S,K>::next
From cppreference.com
constexpr subrange next( std::iter_difference_t<I> n = 1 ) const& requires std::forward_iterator<I>; |
(1) | (since C++20) |
constexpr subrange next( std::iter_difference_t<I> n = 1 ) &&; |
(2) | (since C++20) |
Returns a subrange
whose begin_
is incremented (or decremented if n is negative). The actual increment (or decrement) operation is performed by advance()
.
1) Returns a copy of *this. Equivalent to:
auto tmp = *this;
tmp.advance(n);
return tmp;.
tmp.advance(n);
return tmp;.
2) Returns a
subrange
moved from *this. Equivalent to: advance(n);
return std::move(*this);.
return std::move(*this);.
Contents |
[edit] Parameter
n | - | number of maximal increments of the iterator |
[edit] Return value
As described above.
[edit] Notes
The difference between this function and advance()
is that the latter performs the increment (or decrement) in place.
[edit] Example
Run this code
#include <array> #include <iterator> #include <print> #include <ranges> int main() { std::array arr{1, 2, 3, 4, 5, 6, 7}; std::ranges::subrange sub{std::next(arr.begin(), 2), std::prev(arr.end(), 2)}; std::println("1) sub: {}", sub); std::println("2) sub: {}", sub.next()); std::println("3) sub: {}", sub.next(2)); }
Output:
1) sub: [3, 4, 5] 2) sub: [4, 5] 3) sub: [5]
[edit] See also
obtains a copy of the subrange with its iterator decremented by a given distance (public member function) | |
advances the iterator by given distance (public member function) | |
(C++11) |
increment an iterator (function template) |
(C++20) |
increment an iterator by a given distance or to a bound (algorithm function object) |