std::ranges::subrange<I,S,K>::next
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std::ranges::subrange
subrange::next
constexpr subrange next( std::iter_difference_t <I> n = 1 ) const&
requires std::forward_iterator <I>;
(1)
(since C++20)
requires std::forward_iterator <I>;
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
(public member function) [edit]
subrange with its iterator decremented by a given distance (public member function) [edit]