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调用签名
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template< std::input_iterator I1, std::sentinel_for<I1> S1,
std::input_iterator I2, std::sentinel_for<I2> S2,
class Proj1 = std::identity, class Proj2
= std::identity,
std::indirect_strict_weak_order<
std::projected<I1,
Proj1>,
std::projected<I2,
Proj2>> Comp =
ranges::less >
constexpr bool
lexicographical_compare( I1 first1, S1 last1,
I2 first2, S2 last2,
Comp comp =
{}, Proj1 proj1 = {}, Proj2 proj2 = {} );
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(1) |
(C++20 起) |
template< ranges::input_range R1, ranges::input_range R2,
class Proj1 =
std::identity, class Proj2 = std::identity,
std::indirect_strict_weak_order<
std::projected<ranges::iterator_t<R1>, Proj1>,
std::projected<ranges::iterator_t<R2>, Proj2>> Comp =
ranges::less >
constexpr bool
lexicographical_compare( R1&& r1, R2&& r2, Comp comp = {},
Proj1 proj1 = {}, Proj2 proj2 = {} );
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(2) |
(C++20 起) |
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检查第一范围 [first1,
last1) 是否按字典序小于第二范围 [first2,
last2) 。
1) 用给定的二元比较函数 comp 比较元素。
2) 同 (1) ，但以 r
为源范围，如同以 ranges::begin(r) 为 first 并以 ranges::end(r) 为 last 。
字典序比较是拥有下列属性的操作：
- 按元素逐个比较二个范围。
- 首个不匹配元素定义范围按字典序小于或大于另一个。
- 若一个范围是另一个的前缀，则较短的范围按字典序小于另一个。
- 若二个范围拥有等价的元素与相同长度，则范围按字典序相等。
- 空范围按字典序小于任何非空范围。
- 二个空范围按字典序相等。
此页面上描述的仿函数实体是 niebloid，即：
实际上，它们能以函数对象，或者某些特殊编译器扩展实现。
参数
| first1, last1
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-
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要检验的第一元素范围
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| r1
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-
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要检验的第一元素范围
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| first2, last2
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-
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要检验的第二元素范围
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| r2
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-
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要检验的第二元素范围
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| comp
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-
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应用到投影后元素的比较函数
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| proj1
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-
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应用到第一元素范围的投影
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| proj2
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-
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应用到第二元素范围的投影
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返回值
若第一范围按字典序小于第二范围则为 true 。
复杂度
至多应用 2·min(N1, N2) 次比较与对应的投影，其中 N1 = ranges::distance(first1, last1) 而 N2 = ranges::distance(first2, last2) 。
可能的实现
struct lexicographical_compare_fn {
template<std::input_iterator I1, std::sentinel_for<I1> S1,
std::input_iterator I2, std::sentinel_for<I2> S2,
class Proj1 = std::identity, class Proj2 = std::identity,
std::indirect_strict_weak_order<
std::projected<I1, Proj1>,
std::projected<I2, Proj2>> Comp = ranges::less>
constexpr bool operator()(I1 first1, S1 last1, I2 first2, S2 last2,
Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
for ( ; (first1 != last1) && (first2 != last2); ++first1, (void) ++first2 ) {
if (std::invoke(comp, std::invoke(proj1, *first1), std::invoke(proj2, *first2))) {
return true;
}
if (std::invoke(comp, std::invoke(proj2, *first2), std::invoke(proj1, *first1))) {
return false;
}
}
return (first1 == last1) && (first2 != last2);
}
template< ranges::input_range R1, ranges::input_range R2,
class Proj1 = std::identity, class Proj2 = std::identity,
std::indirect_strict_weak_order<
std::projected<ranges::iterator_t<R1>, Proj1>,
std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less >
constexpr bool operator()(R1&& r1, R2&& r2, Comp comp = {},
Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
return (*this)(ranges::begin(r1), ranges::end(r1),
ranges::begin(r2), ranges::end(r2),
std::ref(comp), std::ref(proj1), std::ref(proj2));
}
};
inline constexpr lexicographical_compare_fn lexicographical_compare;
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示例
#include <algorithm>
#include <iterator>
#include <iostream>
#include <vector>
#include <random>
int main()
{
std::vector<char> v1 {'a', 'b', 'c', 'd'};
std::vector<char> v2 {'a', 'b', 'c', 'd'};
namespace ranges = std::ranges;
std::mt19937 g{std::random_device{}()};
while (!ranges::lexicographical_compare(v1, v2)) {
ranges::copy(v1, std::ostream_iterator<char>(std::cout, " "));
std::cout << ">= ";
ranges::copy(v2, std::ostream_iterator<char>(std::cout, " "));
std::cout << '\n';
ranges::shuffle(v1, g);
ranges::shuffle(v2, g);
}
ranges::copy(v1, std::ostream_iterator<char>(std::cout, " "));
std::cout << "< ";
ranges::copy(v2, std::ostream_iterator<char>(std::cout, " "));
std::cout << '\n';
}
可能的输出：
a b c d >= a b c d
d a b c >= c b d a
b d a c >= a d c b
a c d b < c d a b
参阅
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确定两个元素集合是否是相同的
(niebloid) |
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当一个范围按字典顺序小于另一个范围时，返回 true
(函数模板) |