std::max

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.

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).

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.

Complexity

1-2) Exactly one comparison

3-4) Exactly ilist.size() - 1 comparisons

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); }

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

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

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