Re: Internal move vs. copy in std::vector
Daniel Kr?gler wrote:
On 2012-05-07 02:03, Wil Evers wrote:
I experimented a little bit with gcc-4.7.0, and FWIW, I did run
into a surprise when adding move semantics to a C++-98-style
noncopyable type. This is what I tried:
#include<iostream>
#include<vector>
class elem {
public :
elem() { }
elem(elem&&) { std::cout<< "elem(elem&&)"<< std::endl; }
private :
// Disable copying (C++98 style)
elem(const elem&); // not defined
};
int main()
{
std::vector<elem> vec;
vec.emplace_back();
vec.reserve(vec.capacity() + 1);
return 0;
}
This caused the compiler to complain that elem's copy constructor
is not accessible. (The actual instantiation traceback is quite
interesting, but I'll leave that out for now.) Because elem's move
constructor may throw, the compiler insists on using the copy
constructor - even though it is inaccessible.
To get the code to compile, one has to either decorate the move
constructor with 'nothrow', or the copy constructor with '=
delete'. I wonder how many C++ users will understand which of
these to pick, and when.
Note that this observation is based on a compiler that still does
not completely implement the C++11 standard. As of C++11 access
"violations" will be considered as part of "sfinae" conditions, that
is a private copy constructor of type X *should* have the effect
that std::is_copy_consructible<X>::value evaluates to false.
That's good news to me; thanks for pointing it out. I found out later
that simply removing the private copy constructor declaration also
works, because the compiler will not generate a copy constructor (or
copy assignment operator) for a type with an explicitly declared move
constructor.
In that case, an attempt to call the copy constructor causes gcc-4.7.0
to report that because of the presence of the move constructor, the
copy constructor is implictly declared as deleted. To me, this seems
to be the most elegant approach.
Regards,
- Wil
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