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C++11: Are there reasons why some Regular Types should not have `std::hash` specialised?

With a Regular Type, I mean the definition of Stepanov in Elements of Programming, basically, that there's the concept of equality and that objects which are copies of each other compare equal.

So when you have a Regular Type T, and the equality relation is transitive (a == b && b == c => a == c), you can define a (non-trivial) hash function which is consistent with the definition of equality (a == b => h(a) == h(b)). Always.

But the standard doesn't include many std::hash specialisations. E.g. std::complex doesn't have one, and neither have the containers, with the notable exceptions of vector<bool> and bitset.

So I'm wondering what the design principle is here.

Or, asked differently: Are there reasons not to provide std::hash specialisations for your own types, provided they are regular and equality is transitive?

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Marc Mutz - mmutz Avatar asked Apr 30 '15 13:04

Marc Mutz - mmutz


1 Answers

Yes.

When a type has the following two properties I do not think you should define std::hash:

  • There is no efficient way to consistently create a quality hash that covers all the data used to describe equality.

  • There is no efficient and/or intuitive way to select a consistent subset of data to hash.

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orlp Avatar answered Sep 19 '22 19:09

orlp