std::ranges::find_end
Defined in header <algorithm>
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Call signature |
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template< std::forward_iterator I1, std::sentinel_for<I1> S1, std::forward_iterator I2, std::sentinel_for<I2> S2, |
(1) | (since C++20) |
template< ranges::forward_range R1, ranges::forward_range R2, class Pred = ranges::equal_to, |
(2) | (since C++20) |
[
first2,
last2)
in the range [
first1,
last1)
, after projection with proj1 and proj2 respectively. The projected elements are compared using the binary predicate pred.The function-like entities described on this page are niebloids, that is:
- Explicit template argument lists cannot be specified when calling any of them.
- None of them are visible to argument-dependent lookup.
- When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.
In practice, they may be implemented as function objects, or with special compiler extensions.
Contents |
[edit] Parameters
first1, last1 | - | the range of elements to examine (aka haystack) |
first2, last2 | - | the range of elements to search for (aka needle) |
r1 | - | the range of elements to examine (aka haystack) |
r2 | - | the range of elements to search for (aka needle) |
pred | - | binary predicate to compare the elements |
proj1 | - | projection to apply to the elements in the first range |
proj2 | - | projection to apply to the elements in the second range |
[edit] Return value
[
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}.[edit] Complexity
At most S·(N-S+1) applications of the corresponding predicate and each projection, where S is ranges::distance(first2, last2) and N is ranges::distance(first1, last1) for (1), or S is ranges::distance(r2) and N is ranges::distance(r1) for (2).
[edit] 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.
[edit] Possible implementation
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 {}; |
[edit] Example
#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) ^^^^^
[edit] See also
(C++23)(C++23)(C++23) |
finds the last element satisfying specific criteria (niebloid) |
(C++20)(C++20)(C++20) |
finds the first element satisfying specific criteria (niebloid) |
(C++20) |
searches for any one of a set of elements (niebloid) |
(C++20) |
finds the first two adjacent items that are equal (or satisfy a given predicate) (niebloid) |
(C++20) |
searches for the first occurrence of a range of elements (niebloid) |
(C++20) |
searches for the first occurrence of a number consecutive copies of an element in a range (niebloid) |
finds the last sequence of elements in a certain range (function template) |