std::max

From cppreference.com
< cpp‎ | algorithm
 
 
 
Defined in header <algorithm>
(1)
template< class T >
const T& max( const T& a, const T& b );
(until C++14)
template< class T >
constexpr const T& max( const T& a, const T& b );
(since C++14)
(2)
template< class T, class Compare >
const T& max( const T& a, const T& b, Compare comp );
(until C++14)
template< class T, class Compare >
constexpr const T& max( const T& a, const T& b, Compare comp );
(since C++14)
(3)
template< class T >
T max( std::initializer_list<T> ilist );
(since C++11)
(until C++14)
template< class T >
constexpr T max( std::initializer_list<T> ilist );
(since C++14)
(4)
template< class T, class Compare >
T max( std::initializer_list<T> ilist, Compare comp );
(since C++11)
(until C++14)
template< class T, class Compare >
constexpr T max( std::initializer_list<T> ilist, Compare comp );
(since C++14)

Returns the greater of the given values.

1-2) Returns the greater of a and b.
3-4) Returns the greatest of the values in initializer list ilist.

The (1,3) versions use operator< to compare the values, the (2,4) versions use the given comparison function comp.

Contents

[edit] Parameters

a, b - the values to compare
ilist - initializer list with the values to compare
comp - comparison function object (i.e. an object that satisfies the requirements of Compare) which returns ​true if 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);

The signature does not need to have const &, but the function object must not modify the objects passed to it.
The types Type1 and Type2 must be such that an object of type T can be implicitly converted to both of them. ​

Type requirements
-
T must meet the requirements of LessThanComparable in order to use overloads (1,3).
-
T must meet the requirements of CopyConstructible in order to use overloads (3,4).

[edit] Return value

1-2) The greater of a and b. If they are equivalent, returns a.
3-4) The greatest value in ilist. If several values are equivalent to the greatest, returns the leftmost one.

[edit] Complexity

1-2) Exactly one comparison
3-4) Exactly ilist.size() - 1 comparisons

[edit] Possible implementation

First version
template<class T> 
const T& max(const T& a, const T& b)
{
    return (a < b) ? b : a;
}
Second version
template<class T, class Compare> 
const T& max(const T& a, const T& b, Compare comp)
{
    return (comp(a, b)) ? b : a;
}
Third version
template< class T >
T max( std::initializer_list<T> ilist)
{
    return *std::max_element(ilist.begin(), ilist.end());
}
Fourth version
template< class T, class Compare >
T max( std::initializer_list<T> ilist, Compare comp )
{
    return *std::max_element(ilist.begin(), ilist.end(), comp);
}

[edit] Notes

Capturing the result of std::max by reference if one of the parameters is rvalue produces a dangling reference if that parameter is returned:

int n = 1;
const int& r = std::max(n-1, n+1);
// r is dangling

[edit] Example

#include <algorithm>
#include <iostream>
#include <string>
 
int main()
{
    std::cout << "larger of 1 and 9999: " << std::max(1, 9999) << '\n'
              << "larger of 'a', and 'b': " << std::max('a', 'b') << '\n'
              << "longest of \"foo\", \"bar\", and \"hello\": " <<
                  std::max( { "foo", "bar", "hello" },
                            [](const std::string& s1, const std::string& s2) {
                                 return s1.size() < s2.size();
                             }) << '\n';
}

Output:

larger of 1 and 9999: 9999
larger of 'a', and 'b': b
longest of "foo", "bar", and "hello": hello

[edit] See also

returns the smaller of the given values
(function template)
(C++11)
returns the larger and the smaller of two elements
(function template)
returns the largest element in a range
(function template)