3657. std::hash<std::filesystem::path> is not enabled

Section: 31.12.6 [fs.class.path] Status: C++23 Submitter: Jiang An Opened: 2022-01-02 Last modified: 2023-11-22

Priority: Not Prioritized

View all other issues in [fs.class.path].

View all issues with C++23 status.

Discussion:

The hash support of std::filesystem::path is provided by std::filesystem::hash_value, but the specialization std::hash<std::filesystem::path> is currently disabled. IMO the specialization should be enabled, and its operator() should return the same value as hash_value.

[2022-01-15; Daniel provides wording]

Previous resolution [SUPERSEDED]:

This wording is relative to N4901.

1. Modify 31.12.4 [fs.filesystem.syn], header <filesystem> synopsis, as indicated:

   #include <compare> // see 17.11.1 [compare.syn]
   
   namespace std::filesystem {
     // 31.12.6 [fs.class.path], paths
     class path;
   
     // 31.12.6.8 [fs.path.nonmember], path non-member functions
     void swap(path& lhs, path& rhs) noexcept;
     size_t hash_value(const path& p) noexcept;
   
     […]
   }
   
   // [fs.path.hash], hash support
   namespace std {
     template<class T> struct hash;
     template<> struct hash<filesystem::path>;
   }
   
   

2. Following subclause 31.12.6.8 [fs.path.nonmember], introduce a new subclause [fs.path.hash], as indicated:

   29.12.6.? Hash support [fs.path.hash]

   template<> struct hash<filesystem::path>;
   

   -?- For an object p of type filesystem::path, hash<filesystem::path>()(p) shall evaluate to the same result as hash_value(p).

[2022-01-18; Daniel improves wording based on reflector discussion feedback]

[2022-01-30; Reflector poll]

Set status to Tentatively Ready after five votes in favour during reflector poll.

[2022-02-10 Approved at February 2022 virtual plenary. Status changed: Tentatively Ready → WP.]

Proposed resolution:

This wording is relative to N4901.

1. Modify 31.12.4 [fs.filesystem.syn], header <filesystem> synopsis, as indicated:

   #include <compare> // see 17.11.1 [compare.syn]
   
   namespace std::filesystem {
     // 31.12.6 [fs.class.path], paths
     class path;
   
     // 31.12.6.8 [fs.path.nonmember], path non-member functions
     void swap(path& lhs, path& rhs) noexcept;
     size_t hash_value(const path& p) noexcept;
   
     […]
   }
   
   // [fs.path.hash], hash support
   namespace std {
     template<class T> struct hash;
     template<> struct hash<filesystem::path>;
   }
   
   

2. Following subclause 31.12.6.8 [fs.path.nonmember], introduce a new subclause [fs.path.hash], as indicated:

   29.12.6.? Hash support [fs.path.hash]

   template<> struct hash<filesystem::path>;
   

   -?- For an object p of type filesystem::path, hash<filesystem::path>()(p) evaluates to the same result as filesystem::hash_value(p).