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identity<void>
seems brokenSection: 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 aroundstd::identity
(823, 700(i), 939(i)) is that it was standardized as something very different (an unconditional type mapper) than traditional usage indicated (a function object that should derive fromstd::unary_function)
, as the SGI definition does. This issue could be solved, ifstd::identity
is removed (one proposal of 939(i)), 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 fromunary_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 useidentity_of
instead ofidentity
.
[ 2009-10 Santa Cruz: ]
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?).