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