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;.
Contents |
[edit] Parameters
n | - | number of decrements of the iterator |
[edit] Return value
As described above.
[edit] Notes
The difference between this function and advance()
is that the latter performs the decrement (or increment) in place.
[edit] 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]
[edit] 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 (niebloid) |