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std::ranges::find_first_of
Defined in header <algorithm> |
||
---|---|---|
Call signature | ||
|
(1) | (since C++20) |
|
(2) | (since C++20) |
[
first1
,
last1
)
for any of the elements in the range [
first2
,
last2
)
, after projecting the ranges with proj1
and proj2
respectively. The projected elements are compared using the binary predicate pred
.
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, 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.
Parameters
first1, last1 | - | the range of elements to examine (aka haystack) |
first2, last2 | - | the range of elements to search for (aka needles) |
r1 | - | the range of elements to examine (aka haystack) |
r2 | - | the range of elements to search for (aka needles) |
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 |
Return value
Iterator to the first element in the range [
first1
,
last1
)
that is equal to an element from the range [
first2
,
last2
)
after projection. If no such element is found, an iterator comparing equal to last1
is returned.
Complexity
At most S * N
applications of the predicate and each projection, where
(1) S = ranges::distance(first2, last2)
and N = ranges::distance(first1, last1)
;
(2) S = ranges::distance(r2)
and N = ranges::distance(r1)
.
Possible implementation
|
Example
#include <algorithm>
#include <iostream>
#include <iterator>
int main()
{
namespace rng = std::ranges;
constexpr static auto haystack = {1, 2, 3, 4};
constexpr static auto needles = {0, 3, 4, 3};
constexpr auto found1 = rng::find_first_of(haystack.begin(), haystack.end(),
needles.begin(), needles.end());
static_assert(std::distance(haystack.begin(), found1) == 2);
constexpr auto found2 = rng::find_first_of(haystack, needles);
static_assert(std::distance(haystack.begin(), found2) == 2);
constexpr static auto negatives = {-6, -3, -4, -3};
constexpr auto not_found = rng::find_first_of(haystack, negatives);
static_assert(not_found == haystack.end());
constexpr auto found3 = rng::find_first_of(haystack, negatives,
[](int x, int y) { return x == -y; }); // uses a binary comparator
static_assert(std::distance(haystack.begin(), found3) == 2);
struct P { int x, y; };
constexpr static auto p1 = {P{1, -1}, P{2, -2}, P{3, -3}, P{4, -4}};
constexpr static auto p2 = {P{5, -5}, P{6, -3}, P{7, -5}, P{8, -3}};
// Compare only P::y data members by projecting them:
const auto found4 = rng::find_first_of(p1, p2, {}, &P::y, &P::y);
std::cout << "First equivalent element {" << found4->x << ", " << found4->y
<< "} was found at position " << std::distance(p1.begin(), found4)
<< ".\n";
}
Output:
First equivalent element {3, -3} was found at position 2.
See also
searches for any one of a set of elements (function template) |
|
(C++20)
|
finds the first two adjacent items that are equal (or satisfy a given predicate) (niebloid) |
(C++20)(C++20)(C++20)
|
finds the first element satisfying specific criteria (niebloid) |
(C++20)
|
finds the last sequence of elements in a certain range (niebloid) |
(C++20)
|
searches for a range of elements (niebloid) |
(C++20)
|
searches for a number consecutive copies of an element in a range (niebloid) |
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