Re: [boost] [optional] operator<(optional<T>, T) -- is it wrong?

Subject: Re: [boost] [optional] operator<(optional<T>, T) -- is it wrong?
From: Niall Douglas (s_sourceforge_at_[hidden])
Date: 2014-12-02 16:01:15

On 2 Dec 2014 at 15:38, Howard Hinnant wrote:

> > Actually we need a new std::safe_hash<> I think, one explicitly 
> > prohibited from being a trivial hash. I'd personally like to see that 
> > become the default hash for unordered_map et al, and let the 
> > programmer choose std::hash where safe.
> 
> I´d like to see us enable the programmer to *easily* select hash
> algorithms we haven´t even heard of yet.  E.g. perhaps they are being
> invented tomorrow. 

I didn't raise N3980 because I personally think it's orthogonal to 
the use cases for std::hash (and, for reference, I would be a *huge* 
supporter of N3980).

I'd really like if it were possible to declare that for all 
unordered_map<> used in a namespace X that the default hash used is 
some type Y. One could evilly achieve this using a local namespace 
bind like this:

namespace X {
  template<class Key, class T, class Hash=Y<Key>, ...> using 
unordered_map = std::unordered_map<Key, T, Hash, ...>;
}

... but I'd personally prefer if every user of unordered_map were 
*required* to choose their hash via binding above, and the std 
implementation doesn't default a hash. I guess backwards 
compatibility will probably prevent that being possible sadly, unless 
we deprecate unordered_map for something better in the STL of course.

Niall

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