std::ranges::adjacent_find() algorithm
- od C++20
- Simplified
- Detailed
// (1)
constexpr I
adjacent_find( I first, S last, Pred pred = {}, Proj proj = {} );
// (2)
constexpr ranges::borrowed_iterator_t<R>
adjacent_find( R&& r, Pred pred = {}, Proj proj = {} );
The type of arguments are generic and have the following constraints:
-
I
-std::forward_iterator
-
S
-std::sentinel_for<I>
-
Proj
- (none) -
Pred
:- (1) -
indirect_binary_predicate<projected<I, Proj>, projected<I, Proj>>
- (2) -
indirect_binary_predicate<projected<ranges::iterator_t<R>, Proj>, projected<ranges::iterator_t<R>, Proj>>
(The
std::
namespace was ommitted here for readability) - (1) -
-
(2) -
R
-std::ranges::input_range
The Proj
template argument has a default type of std::identity
for all overloads.
// (1)
template<
std::forward_iterator I,
std::sentinel_for<I> S,
class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<I, Proj>,
std::projected<I, Proj>> Pred = ranges::equal_to
>
constexpr I adjacent_find( I first, S last, Pred pred = {}, Proj proj = {} );
// (2)
template<
ranges::forward_range R,
class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<ranges::iterator_t<R>, Proj>,
std::projected<ranges::iterator_t<R>, Proj>> Pred = ranges::equal_to
>
constexpr ranges::borrowed_iterator_t<R> adjacent_find( R&& r, Pred pred = {}, Proj proj = {} );
Searches the range [first
; last
) for two consecutive equal elements.
-
(1) Elements are compared using
pred
(after projecting with the projectionproj
). -
(2) Same as (1), but uses
r
as the source range, as if usingranges::begin(r)
asfirst
andranges::end(r)
aslast
.
The function-like entities described on this page are niebloids.
Parameters
first last | The range of elements to examine. |
r | The range of elements to examine. |
pred | Predicate to apply to the projected elements. |
proj | Projection to apply to the elements. |
Return value
An iterator to the first of the first pair of identical elements, that is, the first iterator it such that bool(std::invoke(pred, std::invoke(proj1, *it), std::invoke(proj, *(it + 1))))
is true
.
If no such elements are found, an iterator equal to last
is returned.
Complexity
Exactly min((result - first) + 1, (last - first) - 1)
applications of the predicate and projection where result is the return value.
Exceptions
(none)
Possible implementation
adjacent_find(1) and adjacent_find(2)
struct adjacent_find_fn
{
template<std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<I, Proj>,
std::projected<I, Proj>> Pred = ranges::equal_to>
constexpr I operator()(I first, S last, Pred pred = {}, Proj proj = {}) const
{
if (first == last)
return first;
auto next = ranges::next(first);
for (; next != last; ++next, ++first)
if (std::invoke(pred, std::invoke(proj, *first), std::invoke(proj, *next)))
return first;
return next;
}
template<ranges::forward_range R, class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<ranges::iterator_t<R>, Proj>,
std::projected<ranges::iterator_t<R>, Proj>> Pred = ranges::equal_to>
constexpr ranges::borrowed_iterator_t<R>
operator()(R&& r, Pred pred = {}, Proj proj = {}) const
{
return (*this)(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj));
}
};
inline constexpr adjacent_find_fn adjacent_find;
Examples
#include <algorithm>
#include <functional>
#include <iostream>
int main()
{
const auto v = {0, 1, 2, 3, 40, 40, 41, 41, 5}; /*
^^ ^^ */
namespace ranges = std::ranges;
if (auto it = ranges::adjacent_find(v.begin(), v.end()); it == v.end())
std::cout << "No matching adjacent elements\n";
else
std::cout << "The first adjacent pair of equal elements is at ["
<< ranges::distance(v.begin(), it) << "] == " << *it << '\n';
if (auto it = ranges::adjacent_find(v, ranges::greater()); it == v.end())
std::cout << "The entire vector is sorted in ascending order\n";
else
std::cout << "The last element in the non-decreasing subsequence is at ["
<< ranges::distance(v.begin(), it) << "] == " << *it << '\n';
}
The first adjacent pair of equal elements is at [4] == 40
The last element in the non-decreasing subsequence is at [7] == 41
Hover to see the original license.