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std::minmax( ) algorithm

// (1)
template< class T >
constexpr std::pair<const T&, const T&> minmax( const T& a, const T& b );

// (2)
template< class T, class Compare >
constexpr std::pair<const T&, const T&> minmax( const T& a, const T& b,
Compare comp );

// (3)
template< class T >
constexpr std::pair<T, T> minmax( std::initializer_list<T> ilist);

// (4)
template< class T, class Compare >
constexpr std::pair<T, T> minmax( std::initializer_list<T> ilist, Compare comp );

Returns the lowest and the greatest of the given values.

  • (1 - 2) Returns references to the smaller and the greater of a and b.

    • (1) Uses operator<.
    • (2) Uses the given comparison function comp.
  • (3 - 4) Returns the smallest and the greatest of the values in initializer list ilist.

    • (3) Uses operator<.
    • (4) Uses the given comparison function comp.

Parameters

a
b

The values to compare.

ilist

Initializer list with the values to compare.

cmp

Comparison function object (i.e. an object that satisfies the requirements of Compare) which returns true if a is less than b. The signature of the comparison function should be equivalent to the following:

bool cmp(const Type1 &a, const Type2 &b);

While the signature does not need to have const&, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) Type1 and Type2 regardless of value category (thus, Type1& is not allowed, nor is Type1 unless for Type1 a move is equivalent to a copy (od C++11)).

The types Type1 and Type2 must be such that an object of type T can be implicitly converted to both of them.

Type requirements

Return value

  • (1 - 2)

    • If a < b or a == b, returns the result of std::pair<const T&, const T&>(a, b)
    • Otherwise, returns std::pair<const T&, const T&>(b, a).
  • (3 - 4) A pair with the smallest value in ilist as the first element and the greatest as the second.
    If several elements are equivalent to the smallest, the leftmost such element is returned.
    If several elements are equivalent to the largest, the rightmost such element is returned.

Complexity

  • (1 - 2) Exactly one comparison.
  • (3 - 4) At most ilist.size() * 3 / 2 comparisons.

Exceptions

(none)

Possible implementation

minmax (1)
template<class T>
constexpr std::pair<const T&, const T&> minmax(const T& a, const T& b)
{
return (b < a) ? std::pair<const T&, const T&>(b, a)
: std::pair<const T&, const T&>(a, b);
}
minmax (2)
template<class T, class Compare>
constexpr std::pair<const T&, const T&> minmax(const T& a, const T& b, Compare comp)
{
return comp(b, a) ? std::pair<const T&, const T&>(b, a)
: std::pair<const T&, const T&>(a, b);
}
minmax (3)
template<class T>
constexpr std::pair<T, T> minmax(std::initializer_list<T> ilist)
{
auto p = std::minmax_element(ilist.begin(), ilist.end());
return std::pair(*p.first, *p.second);
}
minmax (4)
template<class T, class Compare>
constexpr std::pair<T, T> minmax(std::initializer_list<T> ilist, Compare comp)
{
auto p = std::minmax_element(ilist.begin(), ilist.end(), comp);
return std::pair(*p.first, *p.second);
}

Notes

Undefined Behavior

For overloads (1 - 2), if one of the parameters is a temporary, the reference returned becomes a dangling reference at the end of the full expression that contains the call to minmax:

int n = 1;
auto p = std::minmax(n, n + 1);
int m = p.first; // ok
int x = p.second; // undefined behavior

// Note that structured bindings have the same issue
auto [mm, xx] = std::minmax(n, n + 1);
xx; // undefined behavior

Examples

Main.cpp
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <vector>

int main()
{
std::vector<int> v {3, 1, 4, 1, 5, 9, 2, 6};
std::srand(std::time(0));
std::pair<int, int> bounds = std::minmax(std::rand() % v.size(),
std::rand() % v.size());

std::cout << "v[" << bounds.first << "," << bounds.second << "]: ";
for (int i = bounds.first; i < bounds.second; ++i)
std::cout << v[i] << ' ';
std::cout << '\n';
}
Possible output
v[2,7]: 4 1 5 9 2
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::minmax( ) algorithm

// (1)
template< class T >
constexpr std::pair<const T&, const T&> minmax( const T& a, const T& b );

// (2)
template< class T, class Compare >
constexpr std::pair<const T&, const T&> minmax( const T& a, const T& b,
Compare comp );

// (3)
template< class T >
constexpr std::pair<T, T> minmax( std::initializer_list<T> ilist);

// (4)
template< class T, class Compare >
constexpr std::pair<T, T> minmax( std::initializer_list<T> ilist, Compare comp );

Returns the lowest and the greatest of the given values.

  • (1 - 2) Returns references to the smaller and the greater of a and b.

    • (1) Uses operator<.
    • (2) Uses the given comparison function comp.
  • (3 - 4) Returns the smallest and the greatest of the values in initializer list ilist.

    • (3) Uses operator<.
    • (4) Uses the given comparison function comp.

Parameters

a
b

The values to compare.

ilist

Initializer list with the values to compare.

cmp

Comparison function object (i.e. an object that satisfies the requirements of Compare) which returns true if a is less than b. The signature of the comparison function should be equivalent to the following:

bool cmp(const Type1 &a, const Type2 &b);

While the signature does not need to have const&, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) Type1 and Type2 regardless of value category (thus, Type1& is not allowed, nor is Type1 unless for Type1 a move is equivalent to a copy (od C++11)).

The types Type1 and Type2 must be such that an object of type T can be implicitly converted to both of them.

Type requirements

Return value

  • (1 - 2)

    • If a < b or a == b, returns the result of std::pair<const T&, const T&>(a, b)
    • Otherwise, returns std::pair<const T&, const T&>(b, a).
  • (3 - 4) A pair with the smallest value in ilist as the first element and the greatest as the second.
    If several elements are equivalent to the smallest, the leftmost such element is returned.
    If several elements are equivalent to the largest, the rightmost such element is returned.

Complexity

  • (1 - 2) Exactly one comparison.
  • (3 - 4) At most ilist.size() * 3 / 2 comparisons.

Exceptions

(none)

Possible implementation

minmax (1)
template<class T>
constexpr std::pair<const T&, const T&> minmax(const T& a, const T& b)
{
return (b < a) ? std::pair<const T&, const T&>(b, a)
: std::pair<const T&, const T&>(a, b);
}
minmax (2)
template<class T, class Compare>
constexpr std::pair<const T&, const T&> minmax(const T& a, const T& b, Compare comp)
{
return comp(b, a) ? std::pair<const T&, const T&>(b, a)
: std::pair<const T&, const T&>(a, b);
}
minmax (3)
template<class T>
constexpr std::pair<T, T> minmax(std::initializer_list<T> ilist)
{
auto p = std::minmax_element(ilist.begin(), ilist.end());
return std::pair(*p.first, *p.second);
}
minmax (4)
template<class T, class Compare>
constexpr std::pair<T, T> minmax(std::initializer_list<T> ilist, Compare comp)
{
auto p = std::minmax_element(ilist.begin(), ilist.end(), comp);
return std::pair(*p.first, *p.second);
}

Notes

Undefined Behavior

For overloads (1 - 2), if one of the parameters is a temporary, the reference returned becomes a dangling reference at the end of the full expression that contains the call to minmax:

int n = 1;
auto p = std::minmax(n, n + 1);
int m = p.first; // ok
int x = p.second; // undefined behavior

// Note that structured bindings have the same issue
auto [mm, xx] = std::minmax(n, n + 1);
xx; // undefined behavior

Examples

Main.cpp
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <vector>

int main()
{
std::vector<int> v {3, 1, 4, 1, 5, 9, 2, 6};
std::srand(std::time(0));
std::pair<int, int> bounds = std::minmax(std::rand() % v.size(),
std::rand() % v.size());

std::cout << "v[" << bounds.first << "," << bounds.second << "]: ";
for (int i = bounds.first; i < bounds.second; ++i)
std::cout << v[i] << ' ';
std::cout << '\n';
}
Possible output
v[2,7]: 4 1 5 9 2
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.