deductionguidesforstd::multiset (3) - Linux Manuals

deductionguidesforstd::multiset: deductionguidesforstd::multiset

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 std::iterator_traits<InputIt>::value_type>> (1) (since C++17)
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 = Alloc()) (2) (since C++17)
-> multiset<Key, Comp, Alloc>;
template<class InputIt, class Alloc>
multiset(InputIt, InputIt, Alloc) (3) (since C++17)
-> multiset<typename std::iterator_traits<InputIt>::value_type,
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 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

// 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::multiset<int>
  }