How Can I Specialize `std::hash::operator()` for Custom Types in C ?

Specializing std::hash::operator() for User-Defined Types

To leverage unordered containers with user-defined key types, such as std::unordered_set and std::unordered_map, you typically need to define operator==(Key, Key) and a hash functor. However, it may be preferable to utilize a default hash function for such types, as is the case for built-in types.

Upon investigating various resources, including the C Standard, it becomes apparent that it is possible to specialize std::hash::operator() for user-defined types. The following code snippet exemplifies such a specialization:

namespace std {
  template 
  inline size_t hash<x>::operator()(const X& x) const {
    return hash<int>()(x.id);
  }
}</int></x>

Now, let's address the questions raised:

1. Legality of Specialization

Adding specializations to the std namespace is not only permitted but encouraged. It allows for the extension of standard capabilities to support user-defined types.

2. Compliant Version of std::hash::operator()

The correct syntax for specializing std::hash::operator() is as follows:

namespace std {
  template 
  struct hash<x> {
    size_t operator()(const X& x) const {
      // Your custom hash function implementation
    }
  };
}</x>

3. Portable Solution

The std::hash specialization demonstrated earlier requires C 11 compatibility, which may not be universally supported by compilers. For increased portability, consider using a non-standard namespace, e.g.:

namespace ht {
  template 
  struct hash<x> {
    // Your custom hash function implementation
  };
}</x>

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