Re: being copy constructible
Ivan Novick wrote:
Does the non-const nature of the
copy constructor stop it being copy constructible in the sense of the
standard?
class S
{
public:
S(){}
S(S& a){} //non-const copy constructor
bool operator==(const S & rhs){return true;};
};
I just tried it. A user defined copy constructor whose parameter is
not const... still appears to be called, in preference to any compiler
generated version.
There is no compiler generated version. This is a copy
constructor, and since there is a user defined copy constructor,
the compiler will not generate one.
Therefore I would think that as long as your non-const copy constructor
properly initializes an object it will work fine in a vector or any
other standard container.
The standard says that any class contained in a standard
container must be CopyConstructible. The presense of a copy
constructor is not, in itself, sufficient for the class to be
CopyConstructible. In particular, in order to be
CopyConstructible, it must be possible to copy const objects
(not the case with his copy constructor), the copied object must
be "equivalent" to the original (true for his copy constructor,
since all objects of the type are equivalent), and the address
operator (unary &) must return a T* (if the object is of type T)
or a T const*, according to the const-ness of the object, and
this pointer must contain the address of the object.
However, why would you make it non-const? This implies that the object
being copied could have its state changed, and if this was the case,
then you would violate the contract of the standard container.
It depends. There could be state not participating in
"equivalence" (although logically, I would consider making such
state mutable). The real problem is that he cannot copy const
objects.
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