std::ranges::find_end() algorithm

since C++20

Simplified

// (1)
constexpr ranges::subrange<I1>
    find_end( I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );

// (2)
constexpr ranges::borrowed_subrange_t<R1>
    find_end( R1&& r1, R2&& r2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );

The type of arguments are generic and have the following constraints:

• I1, I2 - std::forward_iterator
• S1, S2 - std::sentinel_for<I1>, std::sentinel_for<I2>
• Pred - (none)
• Proj1, Proj2 - (none)
• (2) - R1, R2 - std::ranges::forward_range

The Proj1 and Proj2 template arguments have a default type of std::identity for all overloads.

Additionally, each overload has the following constraints:

• (1) - indirectly_comparable<I1, I2, Pred, Proj1, Proj2>
• (2) - indirectly_comparable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, Pred, Proj1, Proj2>

(The std:: namespace was ommitted here for readability)

Detailed

// (1)
template<
  std::forward_iterator I1,
  std::sentinel_for<I1> S1,
  std::forward_iterator I2,
  std::sentinel_for<I2> S2,
  class Pred = ranges::equal_to,
  class Proj1 = std::identity,
  class Proj2 = std::identity
>
  requires std::indirectly_comparable<I1, I2, Pred, Proj1, Proj2>
constexpr ranges::subrange<I1>
    find_end( I1 first1, S1 last1, I2 first2, S2 last2,
              Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );

// (2)
template<
  ranges::forward_range R1,
  ranges::forward_range R2,
  class Pred = ranges::equal_to,
  class Proj1 = std::identity,
  class Proj2 = std::identity
>
  requires std::indirectly_comparable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, Pred, Proj1, Proj2>
constexpr ranges::borrowed_subrange_t<R1>
    find_end( R1&& r1, R2&& r2, Pred pred = {},
              Proj1 proj1 = {}, Proj2 proj2 = {} );

Searches for the last occurrence of a sequence in a range.

• (1) Searches for the last occurrence of the sequence [first2; last2) in the range [first1; last1), after projection with proj1 and proj2 respectively. The projected elements are compared using the binary predicate pred.

• (2) Same as (1), but uses r1 as the first source range and r2 as the second source range, as if using ranges::begin(r1) as first1, ranges::end(r1) as last1, ranges::begin(r2) as first2, and ranges::end(r2) as last2.

The function-like entities described on this page are niebloids.

Parameters

first1 last1	The range of elements to examine.
first2 last2	The range of elements to search for.
r1	The range of elements to examine.
r2	The range of elements to search for.
pred	Binary predicate to compare the elements with.
proj1	Projection to apply to the elements in the first range.
proj2	Projection to apply to the elements in the second range.

Return value

• (1) Value of type ranges::subrange<I1> initialized as follows:

  {
    i,
    i + (i == last1 ? 0 : ranges::distance(first2, last2))
  }

  that denotes the last occurrence of the sequence [first2; last2) in range [first1; last1) (after projections with proj1 and proj2).

  If [first2; last2) is empty or if no such sequence is found, the return value is effectively initialized with { last1, last1 }.

• (2) Same as (1), except that the return type is ranges::borrowed_subrange_t<R1>.

Complexity

• (1) Given S as ranges::distance(first2, last2) and N as ranges::distance(first1, last1)
• (2) Given S as ranges::distance(r2) and N as ranges::distance(r1)

At most S * (N - S + 1) applications of predicate and each projection.

Exceptions

(none)

Possible implementation

find_end(1)

struct find_end_fn
{
    template<std::forward_iterator I1, std::sentinel_for<I1> S1,
             std::forward_iterator I2, std::sentinel_for<I2> S2,
             class Pred = ranges::equal_to,
             class Proj1 = std::identity, class Proj2 = std::identity>
    requires std::indirectly_comparable<I1, I2, Pred, Proj1, Proj2>
    constexpr ranges::subrange<I1>
        operator()(I1 first1, S1 last1,
                   I2 first2, S2 last2, Pred pred = {},
                   Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        if (first2 == last2)
        {
            auto last_it = ranges::next(first1, last1);
            return {last_it, last_it};
        }
        auto result = ranges::search(
            std::move(first1), last1, first2, last2, pred, proj1, proj2);

        if (result.empty()) return result;

        for (;;)
        {
            auto new_result = ranges::search(
                std::next(result.begin()), last1, first2, last2, pred, proj1, proj2);
            if (new_result.empty())
                return result;
            else
                result = std::move(new_result);
        }
    }

    template<ranges::forward_range R1, ranges::forward_range R2,
             class Pred = ranges::equal_to,
             class Proj1 = std::identity,
             class Proj2 = std::identity>
    requires std::indirectly_comparable<ranges::iterator_t<R1>,
                                        ranges::iterator_t<R2>,
                                        Pred, Proj1, Proj2>
    constexpr ranges::borrowed_subrange_t<R1>
        operator()(R1&& r1, R2&& r2, Pred pred = {},
                   Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        return (*this)(ranges::begin(r1), ranges::end(r1),
                       ranges::begin(r2), ranges::end(r2),
                       std::move(pred),
                       std::move(proj1), std::move(proj2));
    }
};

inline constexpr find_end_fn find_end {};

Notes

An implementation can improve efficiency of the search if the input iterators model std::bidirectional_iterator by searching from the end towards the begin. Modelling the std::random_access_iterator may improve the comparison speed.
All this however does not change the theoretical complexity of the worst case.

Examples

Main.cpp

#include <algorithm>
#include <array>
#include <cctype>
#include <iostream>
#include <ranges>
#include <string_view>

void print(const auto haystack, const auto needle)
{
    const auto pos = std::distance(haystack.begin(), needle.begin());
    std::cout << "In \"";
    for (const auto c : haystack) { std::cout << c; }
    std::cout << "\" found \"";
    for (const auto c : needle) { std::cout << c; }
    std::cout << "\" at position [" << pos << ".." << pos + needle.size() << ")\n"
        << std::string(4 + pos, ' ') << std::string(needle.size(), '^') << '\n';
}

int main()
{
    using namespace std::literals;
    constexpr auto secret{"password password word..."sv};
    constexpr auto wanted{"password"sv};

    constexpr auto found1 = std::ranges::find_end(
        secret.cbegin(), secret.cend(), wanted.cbegin(), wanted.cend());
    print(secret, found1);

    constexpr auto found2 = std::ranges::find_end(secret, "word"sv);
    print(secret, found2);

    const auto found3 = std::ranges::find_end(secret, "ORD"sv,
        [](const char x, const char y) { // uses a binary predicate
            return std::tolower(x) == std::tolower(y);
        });
    print(secret, found3);

    const auto found4 = std::ranges::find_end(secret, "SWORD"sv, {}, {},
        [](char c) { return std::tolower(c); }); // projects the 2nd range
    print(secret, found4);

    static_assert(std::ranges::find_end(secret, "PASS"sv).empty()); // => not found
}

Output

In "password password word..." found "password" at position [9..17)
             ^^^^^^^^
In "password password word..." found "word" at position [18..22)
                      ^^^^
In "password password word..." found "ord" at position [19..22)
                       ^^^
In "password password word..." found "sword" at position [12..17)
                ^^^^^