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deductiongu...td::multiset(3) C++ Standard Libarydeductiongu...td::multiset(3)

NAME
       deductionguidesforstd::multiset - deductionguidesforstd::multiset

Synopsis
	  Defined in header <set>
	  template<class InputIt,

	  class	Comp = std::less<typename
	  std::iterator_traits<InputIt>::value_type>,
	  class		   Alloc	   =	       std::allocator<typename
       (1) (since C++17)
	  std::iterator_traits<InputIt>::value_type>>
	  multiset(InputIt, InputIt, Comp = Comp(), Alloc = Alloc())

	  -> multiset<typename std::iterator_traits<InputIt>::value_type,
	  Comp,	Alloc>;
	  template<class Key, class Comp = std::less<Key>, class Alloc =
	  std::allocator<Key>>

	  multiset(std::initializer_list<Key>,	Comp   =   Comp(),   Alloc   =
       (2) (since C++17)
	  Alloc())

	  -> multiset<Key, Comp, Alloc>;
	  template<class InputIt, class	Alloc>

	  multiset(InputIt, InputIt, Alloc)
	  ->	multiset<typename   std::iterator_traits<InputIt>::value_type,
       (3) (since C++17)

	  std::less<typename std::iterator_traits<InputIt>::value_type>,
	  Alloc>;
	  template<class Key, class Alloc>

	  multiset(std::initializer_list<Key>,				Alloc)
       (4) (since C++17)

	  -> multiset<Key, std::less<Key>, Alloc>;

	  These	 deduction guides are provided for multiset to allow deduction
       from an iterator
	  range	(overloads (1,3)) and std::initializer_list (overloads (2,4)).
       These overloads
	  participate in overload resolution only if InputIt satisfies Legacy-
       InputIterator,
	  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 inte-
       gral types do not
	  qualify as input iterators. Likewise,	the extent to which it	deter-
       mines 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
       // Run this code

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

http://cppreference.com		  2022.07.31	 deductiongu...td::multiset(3)

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