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Home/ Questions/Q 9012553
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T03:01:49+00:00 2026-06-16T03:01:49+00:00

I have a class C, which has a string* ps private data member. Now,

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I have a class C, which has a string* ps private data member.
Now, I’d like to have an unordered_map<C, int> for which I need a custom hash function.

According to the c++ reference, I can do that like

namespace std {
  template<>
  class hash<C> {
  public:
    size_t operator()(const C &c) const
    {
      return std::hash<std::string>()(*c.ps);
    }
  };
}

The problem is that I can’t seem to make operator() and C friends so that I could access ps.

I have tried this:

class C;
template<>
class std::hash<C>;
class C{
  //...
  friend std::hash<C>::operator ()(const C&) const; // error: Incomplete type 
};
// define hash<C> here.

but it says that Incomplete type … in nested name specifier …

I can’t turn around the definitions either, because if class C is defined later, the hash<C> has no way to know about ps.

What am I doing wrong here? How can this situation be fixed without making ps public?

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  1. Editorial Team
    Editorial Team
    2026-06-16T03:01:50+00:00Added an answer on June 16, 2026 at 3:01 am

    Try this:

    class C;
    namespace std {
      template<>
      struct hash<C> {
      public:
        size_t operator()(const C &c) const; // don't define yet
      };
    }
    class C{
      //...
      friend size_t std::hash<C>::operator ()(const C&) const;
    };
    namespace std {
      template<>
      size_t hash<C>::operator()(const C &c) const {
        return std::hash<std::string>()(*c.ps);
      }
    }
    

    Or this:

    class C;
    template<>
    struct std::hash<C>;
    class C{
      friend struct std::hash<C>; // friend the class, not the member function
    };
    

    (I haven’t compiled so there might be a syntax error)

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