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Section: 99 [func.referenceclosure.cons] Status: CD1 Submitter: Lawrence Crowl Opened: 2008-06-02 Last modified: 2016-01-28
Priority: Not Prioritized
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Discussion:
The std::reference_closure type has a deleted copy assignment operator under the theory that references cannot be assigned, and hence the assignment of its reference member must necessarily be ill-formed.
However, other types, notably std::reference_wrapper and std::function provide for the "copying of references", and thus the current definition of std::reference_closure seems unnecessarily restrictive. In particular, it should be possible to write generic functions using both std::function and std::reference_closure, but this generality is much harder when one such type does not support assignment.
The definition of reference_closure does not necessarily imply direct implementation via reference types. Indeed, the reference_closure is best implemented via a frame pointer, for which there is no standard type.
The semantics of assignment are effectively obtained by use of the default destructor and default copy assignment operator via
x.~reference_closure(); new (x) reference_closure(y);
So the copy assignment operator generates no significant real burden to the implementation.
Proposed resolution:
In [func.referenceclosure] Class template reference_closure, replace the =delete in the copy assignment operator in the synopsis with =default.
template<class R , class... ArgTypes >
class reference_closure<R (ArgTypes...)> {
public:
...
reference_closure& operator=(const reference_closure&) = delete default;
...
In 99 [func.referenceclosure.cons] Construct, copy, destroy, add the member function description
reference_closure& operator=(const reference_closure& f)Postcondition: *this is a copy of f.
Returns: *this.