std::find_first_of
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Defined in header <algorithm>
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template< class ForwardIt1, class ForwardIt2 > ForwardIt1 find_first_of( ForwardIt1 first, ForwardIt1 last, |
(1) | (C + + 11時まで) (C + + 11以来) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > ForwardIt1 find_first_of( ForwardIt1 first, ForwardIt1 last, |
(2) | (C + + 11時まで) (C + + 11以来) |
[first, last)
の要素のいずれかのレンジ[s_first, s_last)
を検索します。最初のバージョンは、要素を比較するoperator==
使用して、2番目のバージョンは、指定されたバイナリ述語p
を使用しています. [first, last)
for any of the elements in the range [s_first, s_last)
. The first version uses operator==
to compare the elements, the second version uses the given binary predicate p
. You can help to correct and verify the translation. Click here for instructions.
目次 |
[編集] パラメータ
first, last | - | 検討する要素の範囲
Original: the range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
s_first, s_last | - | 検索する要素の範囲
Original: the range of elements to search for The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
p | - | binary predicate which returns true if the elements should be treated as equal. The signature of the predicate function should be equivalent to the following: bool pred(const Type1 &a, const Type2 &b); The signature does not need to have const &, but the function must not modify the objects passed to it. |
Type requirements | ||
-InputIt must meet the requirements of InputIterator .
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-ForwardIt1 must meet the requirements of ForwardIterator .
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-ForwardIt2 must meet the requirements of ForwardIterator .
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[編集] 値を返します
[first, last)
から要素に等しい範囲[s_first; s_last)
内の最初の要素へのイテレータ。このような要素が見つからない場合、last
が返されます。.[first, last)
that is equal to an element from the range [s_first; s_last)
. If no such element is found, last
is returned.You can help to correct and verify the translation. Click here for instructions.
[編集] 複雑
(S*N)
の比較ではどこS = distance(s_first, s_last)とN = distance(first, last).(S*N)
comparisons where S = distance(s_first, s_last) and N = distance(first, last).You can help to correct and verify the translation. Click here for instructions.
[編集] 可能な実装
First version |
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template<class InputIt, class ForwardIt> InputIt find_first_of(InputIt first, InputIt last, ForwardIt s_first, ForwardIt s_last) { for (; first != last; ++first) { for (ForwardIt it = s_first; it != s_last; ++it) { if (*first == *it) { return first; } } } return last; } |
Second version |
template<class InputIt, class ForwardIt, class BinaryPredicate> InputIt find_first_of(InputIt first, InputIt last, ForwardIt s_first, ForwardIt s_last, BinaryPredicate p) { for (; first != last; ++first) { for (ForwardIt it = s_first; it != s_last; ++it) { if (p(*first, *it)) { return first; } } } return last; } |
[編集] 例
You can help to correct and verify the translation. Click here for instructions.
#include <algorithm> #include <iostream> #include <vector> int main() { std::vector<int> v{0, 2, 3, 25, 5}; std::vector<int> t{3, 19, 10, 2}; auto result = std::find_first_of(v.begin(), v.end(), t.begin(), t.end()); if (result == v.end()) { std::cout << "no elements of v were equal to 3, 19, 10 or 2\n"; } else { std::cout << "found a match at " << std::distance(v.begin(), result) << "\n"; } }
Output:
found a match at 1
[編集] も参照してください
(C++11) |
特定の条件を満たす最初の要素を検索します Original: finds the first element satisfying specific criteria The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (関数テンプレート) |