deduction guides for std::map

From cppreference.com
< cpplrm; | containerlrm; | map

Defined in header <map>
template <class InputIt,

class Comp = std::less<iter_key_t<InputIt>>,
class Alloc = std::allocator<iter_to_alloc_t<InputIt>>>
map(InputIt, InputIt, Comp = Comp(), Alloc = Alloc())

-> map<iter_key_t<InputIt>, iter_val_t<InputIt>, Comp, Alloc>;
(1) (since C++17)
template<class Key,

class T,
class Comp = std::less<Key>,
class Alloc = std::allocator<std::pair<const Key, T>>>
map(std::initializer_list<std::pair<const Key, T>>, Comp = Comp(), Alloc = Alloc())

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

map(InputIt, InputIt, Alloc)
-> map<iter_key_t<InputIt>, iter_val_t<InputIt>,

std::less<iter_key_t<InputIt>>, Alloc>;
(3) (since C++17)
template<class Key, class T, class Allocator>

map(std::initializer_list<std::pair<const Key, T>>, Allocator)

-> map<Key, T, std::less<Key>, Allocator>;
(4) (since C++17)

where the type aliases iter_key_t, iter_val_t, iter_to_alloc_t are defined as if as follows

template<class InputIt>

using iter_key_t = std::remove_const_t<

typename std::iterator_traits<InputIt>::value_type::first_type>;
(exposition only)
template<class InputIt>
using iter_val_t = typename std::iterator_traits<InputIt>::value_type::second_type;
(exposition only)
template<class InputIt>

using iter_to_alloc_t = std::pair<
std::add_const_t<typename std::iterator_traits<InputIt>::value_type::first_type>,
typename std::iterator_traits<InputIt>::value_type::second_type

>
(exposition only)

This deduction guide is provided for map 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 InputIterator, Alloc satisfies Allocator, and Comp does not satisfy Allocator.

Note: the extent to which the library determines that a type does not satisfy InputIterator 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 <map>
int main() {
// std::map m1 = {{"foo", 1}, {"bar", 2}}; // Error: braced-init-list has no type;
                                           // cannot deduce pair<const Key, T> from
                                           // {"foo", 1} or {"bar", 2}

   std::map m1 = std::initializer_list<
                        std::pair<char const* const, int>>({{"foo", 2}, {"bar", 3}}); // guide #2
   std::map m2(m1.begin(), m1.end()); // guide #1
}