Przejdź do głównej zawartości

std::ranges::minmax_element() algorithm

// (1)
constexpr minmax_element_result<I>
minmax_element( I first, S last, Comp comp = {}, Proj proj = {} );

// (2)
constexpr minmax_element_result<ranges::borrowed_iterator_t<R>>
minmax_element( R&& r, Comp comp = {}, Proj proj = {} );

The type of arguments are generic and have the following constraints:

  • I - std::forward_iterator
  • S - std::sentinel_for<I>
  • R - std::ranges::forward_range
  • Comp:
    • (1) - std::indirect_strict_weak_order<std::projected<I, Proj>>
    • (2) - std::indirect_strict_weak_order<std::projected<ranges::iterator_t<R>, Proj>>
  • Proj - (none)

The Proj and Comp template arguments have the following default types: std::identity, ranges::less for all overloads.

With the helper types defined as follows:

template< class I >
using minmax_element_result = ranges::min_max_result<I>;
  • (1) Finds the smallest and largest element in the range [first; last).

  • (2) Same as (1), but uses r as the source range, as if using ranges::begin(r) as first and ranges::end(r) as last.

The function-like entities described on this page are niebloids.

Parameters

first
last

The range to find the smallest and largest value in.

r

The range to find the smallest and largest value in.

comp

Comparison to apply to the projected elements.

proj

Projection to apply to the elements.

Return value

An object consisting of an iterator to the smallest element as the first element and an iterator to the greatest element as the second.

Returns { first, first } if the range is empty.

If several elements are equivalent to the smallest element, the iterator to the first such element is returned.
If several elements are equivalent to the largest element, the iterator to the last such element is returned.

Complexity

Given N as ranges::distance(first, last):

At most max(floor((3 / 2) * (N − 1)), 0) applications of the comparison and twice as many applications of the projection.

Exceptions

(none)

Possible implementation

minmax_element(1) and minmax_element(2)
struct minmax_element_fn
{
template<std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity,
std::indirect_strict_weak_order<std::projected<I, Proj>> Comp = ranges::less>
constexpr ranges::minmax_element_result<I>
operator()(I first, S last, Comp comp = {}, Proj proj = {}) const
{
auto min = first, max = first;

if (first == last || ++first == last) {
return {min, max};
}

if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *min))) {
min = first;
} else {
max = first;
}

while (++first != last) {
auto i = first;
if (++first == last) {
if (std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *min))) {
min = i;
}
else if (!(std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *max)))) {
max = i;
}
break;
} else {
if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *i))) {
if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *min))) {
min = first;
}
if (!(std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *max)))) {
max = i;
}
} else {
if (std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *min))) {
min = i;
}
if (!(std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *max)))) {
max = first;
}
}
}
}
return {min, max};
}

template<ranges::forward_range R, class Proj = std::identity,
std::indirect_strict_weak_order<
std::projected<ranges::iterator_t<R>, Proj>> Comp = ranges::less>
constexpr ranges::minmax_element_result<ranges::borrowed_iterator_t<R>>
operator()(R&& r, Comp comp = {}, Proj proj = {}) const
{
return (*this)(ranges::begin(r), ranges::end(r), std::ref(comp), std::ref(proj));
}
};

inline constexpr minmax_element_fn minmax_element;

Examples

Main.cpp
#include <algorithm>
#include <iostream>
#include <iterator>
namespace rng = std::ranges;

int main()
{
const auto v = {3, 9, 1, 4, 1, 2, 5, 9};
const auto [min, max] = rng::minmax_element(v);
std::cout
<< "min = " << *min << ", at [" << rng::distance(v.begin(), min) << "]\n"
<< "max = " << *max << ", at [" << rng::distance(v.begin(), max) << "]\n";
}
Output
min = 1, at [2]
max = 9, at [7]
This article originates from this CppReference page. It was likely altered for improvements or editors' preference. Click "Edit this page" to see all changes made to this document.
Hover to see the original license.

std::ranges::minmax_element() algorithm

// (1)
constexpr minmax_element_result<I>
minmax_element( I first, S last, Comp comp = {}, Proj proj = {} );

// (2)
constexpr minmax_element_result<ranges::borrowed_iterator_t<R>>
minmax_element( R&& r, Comp comp = {}, Proj proj = {} );

The type of arguments are generic and have the following constraints:

  • I - std::forward_iterator
  • S - std::sentinel_for<I>
  • R - std::ranges::forward_range
  • Comp:
    • (1) - std::indirect_strict_weak_order<std::projected<I, Proj>>
    • (2) - std::indirect_strict_weak_order<std::projected<ranges::iterator_t<R>, Proj>>
  • Proj - (none)

The Proj and Comp template arguments have the following default types: std::identity, ranges::less for all overloads.

With the helper types defined as follows:

template< class I >
using minmax_element_result = ranges::min_max_result<I>;
  • (1) Finds the smallest and largest element in the range [first; last).

  • (2) Same as (1), but uses r as the source range, as if using ranges::begin(r) as first and ranges::end(r) as last.

The function-like entities described on this page are niebloids.

Parameters

first
last

The range to find the smallest and largest value in.

r

The range to find the smallest and largest value in.

comp

Comparison to apply to the projected elements.

proj

Projection to apply to the elements.

Return value

An object consisting of an iterator to the smallest element as the first element and an iterator to the greatest element as the second.

Returns { first, first } if the range is empty.

If several elements are equivalent to the smallest element, the iterator to the first such element is returned.
If several elements are equivalent to the largest element, the iterator to the last such element is returned.

Complexity

Given N as ranges::distance(first, last):

At most max(floor((3 / 2) * (N − 1)), 0) applications of the comparison and twice as many applications of the projection.

Exceptions

(none)

Possible implementation

minmax_element(1) and minmax_element(2)
struct minmax_element_fn
{
template<std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity,
std::indirect_strict_weak_order<std::projected<I, Proj>> Comp = ranges::less>
constexpr ranges::minmax_element_result<I>
operator()(I first, S last, Comp comp = {}, Proj proj = {}) const
{
auto min = first, max = first;

if (first == last || ++first == last) {
return {min, max};
}

if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *min))) {
min = first;
} else {
max = first;
}

while (++first != last) {
auto i = first;
if (++first == last) {
if (std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *min))) {
min = i;
}
else if (!(std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *max)))) {
max = i;
}
break;
} else {
if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *i))) {
if (std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *min))) {
min = first;
}
if (!(std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *max)))) {
max = i;
}
} else {
if (std::invoke(comp, std::invoke(proj, *i), std::invoke(proj, *min))) {
min = i;
}
if (!(std::invoke(comp, std::invoke(proj, *first), std::invoke(proj, *max)))) {
max = first;
}
}
}
}
return {min, max};
}

template<ranges::forward_range R, class Proj = std::identity,
std::indirect_strict_weak_order<
std::projected<ranges::iterator_t<R>, Proj>> Comp = ranges::less>
constexpr ranges::minmax_element_result<ranges::borrowed_iterator_t<R>>
operator()(R&& r, Comp comp = {}, Proj proj = {}) const
{
return (*this)(ranges::begin(r), ranges::end(r), std::ref(comp), std::ref(proj));
}
};

inline constexpr minmax_element_fn minmax_element;

Examples

Main.cpp
#include <algorithm>
#include <iostream>
#include <iterator>
namespace rng = std::ranges;

int main()
{
const auto v = {3, 9, 1, 4, 1, 2, 5, 9};
const auto [min, max] = rng::minmax_element(v);
std::cout
<< "min = " << *min << ", at [" << rng::distance(v.begin(), min) << "]\n"
<< "max = " << *max << ", at [" << rng::distance(v.begin(), max) << "]\n";
}
Output
min = 1, at [2]
max = 9, at [7]
This article originates from this CppReference page. It was likely altered for improvements or editors' preference. Click "Edit this page" to see all changes made to this document.
Hover to see the original license.