deduction guides for std::set

From cppreference.com
< cpp‎ | container‎ | set

Defined in header <set>
template<class InputIt,

         class Comp = std::less<typename std::iterator_traits<InputIt>::value_type>,
         class Alloc = std::allocator<typename std::iterator_traits<InputIt>::value_type>>
set(InputIt, InputIt, Comp = Comp(), Alloc = Alloc())

  -> set<typename std::iterator_traits<InputIt>::value_type, Comp, Alloc>;
(1) (since C++17)
template<class Key, class Comp = std::less<Key>, class Alloc = std::allocator<Key>>

set(std::initializer_list<Key>, Comp = Comp(), Alloc = Alloc())

  -> set<Key, Comp, Alloc>;
(2) (since C++17)
template<class InputIt, class Alloc>

set(InputIt, InputIt, Alloc)
  -> set<typename std::iterator_traits<InputIt>::value_type,

           std::less<typename std::iterator_traits<InputIt>::value_type>, Alloc>;
(3) (since C++17)
template<class Key, class Alloc>

set(std::initializer_list<Key>, Alloc)

  -> set<Key, std::less<Key>, Alloc>;
(4) (since C++17)

These deduction guides are provided for set to allow deduction from an iterator range (overloads (1,3)) and std::initializer_list (overloads (2,4)). These overloads only participate in overload resolution if InputIt satisfies LegacyInputIterator, Alloc satisfies Allocator, and Comp does not satisfy Allocator.

Note: the extent to which the library determines that a type does not satisfy LegacyInputIterator is unspecified, except that as a minimum integral types do not qualify as input iterators. Likewise, the extent to which it determines that a type does not satisfy Allocator is unspecified, except that as a minimum the member type Alloc::value_type must exist and the expression std::declval<Alloc&>().allocate(std::size_t{}) must be well-formed when treated as an unevaluated operand.

Example

#include <set>
int main() {
   std::set s = {1,2,3,4}; // guide #2 deduces std::set<int>
   std::set s2(s.begin(), s.end()); // guide #1 deduces std::set<int>
}