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位置:首页 > C++ 参考手册 >算法库 >有制约算法 (C++20 起) > std::ranges::all_of, std::ranges::any_of, std::ranges::none_of
std::ranges::all_of, std::ranges::any_of, std::ranges::none_of
class Proj = std::identity,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred >
std::indirect_unary_predicate<
std::projected<ranges::iterator_t<R>,Proj>> Pred >
class Proj = std::identity,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred >
std::indirect_unary_predicate<
std::projected<ranges::iterator_t<R>,Proj>> Pred >
class Proj = std::identity,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred >
std::indirect_unary_predicate<
std::projected<ranges::iterator_t<R>,Proj>> Pred >
pred
是否对范围 [first, last)
中的所有元素(以投影 proj
投影后)返回 true 。pred
是否对范围 [first, last)
中至少一个元素(以投影 proj
投影后)返回 true 。pred
是否不对范围 [first, last)
中任何元素(以投影 proj
投影后)返回 true 。r
为源范围,如同以 ranges::begin(r) 为 first
并以 ranges::end(r) 为 last
。此页面上描述的仿函数实体是 niebloids ,即:
实际上,它们能以函数对象,或以某些特殊编译器扩展实现。
参数
first, last | - | 要检验的元素范围 |
r | - | 要检验的元素范围 |
pred | - | 应用到投影后元素的谓词 |
proj | - | 应用到元素的投影 |
返回值
i
成立则为 true ,否则为 false 。若范围为空则返回 true 。i
成立则为 true ,否则为 false 。若范围为空则返回 false 。i
成立则为 true ,否则为 false 。若范围为空则返回 true 。复杂度
至多应用 last - first
次谓词和投影。
可能的实现
版本一 |
---|
struct all_of_fn { template< std::input_iterator I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_unary_predicate<std::projected<I, Proj>> Pred > constexpr bool operator()( I first, S last, Pred pred, Proj proj = {} ) const { return ranges::find_if_not(first, last, std::ref(pred), std::ref(proj)) == last; } template< ranges::input_range R, class Proj = std::identity, std::indirect_unary_predicate< std::projected<ranges::iterator_t<R>,Proj>> Pred > constexpr bool operator()( R&& r, Pred pred, Proj proj = {} ) const { return operator()(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj)); } }; inline constexpr all_of_fn all_of; |
版本二 |
struct any_of_fn { template< std::input_iterator I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_unary_predicate<std::projected<I, Proj>> Pred > constexpr bool operator()( I first, S last, Pred pred, Proj proj = {} ) const { return ranges::find_if(first, last, std::ref(pred), std::ref(proj)) != last; } template< ranges::input_range R, class Proj = std::identity, std::indirect_unary_predicate< std::projected<ranges::iterator_t<R>,Proj>> Pred > constexpr bool operator()( R&& r, Pred pred, Proj proj = {} ) const { return operator()(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj)); } }; inline constexpr any_of_fn any_of; |
版本三 |
struct none_of_fn { template< std::input_iterator I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_unary_predicate<std::projected<I, Proj>> Pred > constexpr bool operator()( I first, S last, Pred pred, Proj proj = {} ) const { return ranges::find_if(first, last, std::ref(pred), std::ref(proj)) == last; } template< ranges::input_range R, class Proj = std::identity, std::indirect_unary_predicate< std::projected<ranges::iterator_t<R>,Proj>> Pred > constexpr bool operator()( R&& r, Pred pred, Proj proj = {} ) const { return operator()(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj)); } }; inline constexpr none_of_fn none_of; |
示例
#include <vector> #include <numeric> #include <algorithm> #include <iterator> #include <iostream> #include <functional> namespace ranges = std::ranges; int main() { std::vector<int> v(10, 2); std::partial_sum(v.cbegin(), v.cend(), v.begin()); std::cout << "Among the numbers: "; ranges::copy(v, std::ostream_iterator<int>(std::cout, " ")); std::cout << '\n'; if (ranges::all_of(v.cbegin(), v.cend(), [](int i){ return i % 2 == 0; })) { std::cout << "All numbers are even\n"; } if (ranges::none_of(v, std::bind(std::modulus<int>(), std::placeholders::_1, 2))) { std::cout << "None of them are odd\n"; } auto DivisibleBy = [](int d) { return [d](int m) { return m % d == 0; }; }; if (ranges::any_of(v, DivisibleBy(7))) { std::cout << "At least one number is divisible by 7\n"; } }
输出:
Among the numbers: 2 4 6 8 10 12 14 16 18 20 All numbers are even None of them are odd At least one number is divisible by 7
参阅
(C++11)(C++11)(C++11) |
检查谓词是否对范围中所有、任一或无元素为 true (函数模板) |