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Section: 22.2.4 [forward] Status: Resolved Submitter: Walter Brown Opened: 2008-04-09 Last modified: 2016-01-28
Priority: Not Prioritized
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Discussion:
N2588 seems to have added an operator() member function to the identity<> helper in 22.2.4 [forward]. I believe this change makes it no longer possible to instantiate identity<void>, as it would require forming a reference-to-void type as this operator()'s parameter type.
Suggested resolution: Specialize identity<void> so as not to require the member function's presence.
[ Sophia Antipolis: ]
Jens: suggests to add a requires clause to avoid specializing on void.
Alisdair: also consider cv-qualified void.
Alberto provided proposed wording.
[ 2009-07-30 Daniel reopens: ]
This issue became closed, because the ReferentType requirement fixed the problem - this is no longer the case. In retrospective it seems to be that the root of current issues around std::identity (823, 700, 939) is that it was standardized as something very different (an unconditional type mapper) than traditional usage indicated (a function object that should derive from std::unary_function), as the SGI definition does. This issue could be solved, if std::identity is removed (one proposal of 939), but until this has been decided, this issue should remain open. An alternative for removing it, would be, to do the following:
Let identity stay as a real function object, which would now properly derive from unary_function:
template <class T> struct identity : unary_function<T, T> { const T& operator()(const T&) const; };Invent (if needed) a generic type wrapper (corresponding to concept IdentityOf), e.g. identity_of, and move it's prototype description back to 22.2.4 [forward]:
template <class T> struct identity_of { typedef T type; };and adapt the std::forward signature to use identity_of instead of identity.
[ 2009-10 Santa Cruz: ]
Mark as
NAD EditorialResolved, fixed by 939.
Proposed resolution:
Change definition of identity in 22.2.4 [forward], paragraph 2, to:
template <class T> struct identity { typedef T type; requires ReferentType<T> const T& operator()(const T& x) const; };
...
requires ReferentType<T> const T& operator()(const T& x) const;
Rationale:
The point here is to able to write T& given T and ReferentType is precisely the concept that guarantees so, according to N2677 (Foundational concepts). Because of this, it seems preferable than an explicit check for cv void using SameType/remove_cv as it was suggested in Sophia. In particular, Daniel remarked that there may be types other than cv void which aren't referent types (int[], perhaps?).