defining hash function for nested custom type

Question

Given a nested template class structure like the following:

template<class T> class A
{
  ...
public:
  ...
  class B;
};

I am wanting the std::hash function to work on the nested class so that it can be put in things like unsorted_map and unsorted_set, so I defined the nested class as:

template<class T> class A<T>::B 
{
  public:
    bool operator==(const A<T>::B &) const;
  ...
  friend struct std::hash<A<T>::B>;
};

And then tried adding a specialized std::hash structure for this type as follows:

namespace std
{
  template<class T> struct hash<A<T>::B>
  {
    bool operator()(const A<T>::B &x) const
    {
       ...
    }
  };
}

But the compiler complains vehemently when I try to define the custom std::hash function object like this.

The error message I get is unhelpful, as follows:

xyz.cc:17:38: error: type/value mismatch at argument 1 in template parameter list for 'template<class _Tp> struct std::hash'
 template<class T> struct hash<A<T>::B>
                                      ^
xyz.cc:17:38: error:   expected a type, got 'A<T>::B'

I'm not sure how else I am supposed to express it, however.

Why is this wrong, and what must I do to to fix it?

I tried `struct hash::B>`, and `struct hash::B>`, which both give the error `template argument 1 is invalid`, and `struct hash::B>` gives the error `template parameters not used in partial specialization`. — markt1964, Aug 20 '15 at 05:43
From what I gather, the only out is define `B` outside of `A`. Either that, or define a specialization for `hash` for each `T` you plan to pass to `A`. — jxh, Aug 20 '15 at 06:20

score 1 · Accepted Answer · answered Aug 21 '15 at 02:07

1

You cannot.

Dependant types cannot be pattern matched against, as the general case requires inverting an arbitrary Turing-complete algorithm.

The easiest way is to make B an independant template, then typedef into A.

answered Aug 21 '15 at 02:07

Yakk - Adam Nevraumont

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defining hash function for nested custom type

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