Re: Do I need a singleton here? RESOLVED

"Jim Langston" <>
Thu, 21 Oct 2010 22:00:44 -0700
"Luc Danton" <> wrote in message

On 22/10/2010 06:11, Jim Langston wrote:

"Bo Persson" <> wrote in message

Jim Langston wrote:

"Jim Langston" <> wrote in message

I am developing a font class for my opengl graphics library and I
came across a quandry. I have a font which is simply a value for
font id and wrapper code to load the font and kill it. Works
fine, then I notice I don't have a virtual destructor, nor any
destructor at all. So I throw in a virtual destructor and call
the code to unload the glfont. I run the program, now no fonts
are visible. The problem is that my fonts are stored in a map by
font name and
the font, and of course std::maps along with most containers use

I used Microsoft Express 2010 C++0x utilization of std::move for
I had a problem implementing this because when I first created a
move constructor I was still using assignment in my code. I
disbled the assignment operator and fixed my code and it works as
advertised. I had to change the elegant line of:
     world.fonts[L"Normal"] = jmlGL::jglFont( hDC, L"Courier New",
24 ); to the no where near as elegant line of:
     world.fonts.insert(std::pair<std::wstring, jmlGL::jglFont>(
L"Normal", jmlGL::jglFont( hDC, L"Courier New", 24 ) ) );

my program worked as expected and I am able to use non-copyable
objects in a container.

Wouldn't another soultion be to implement a move assignment operator
for foo?

foo& operator=(foo&&);

That would let you assign a temporary (rvalue only) to the mapped

I was thinking about this, but if every time I did an assignment I moved
ownership I wouldn't like that. Java does that and it bugs the heck out
of me. As I understand it if I have operator=(&&) then the compiler
will use this instead of operator= (although I might be mistaken in
that). I just don't like the idea of accidently changing ownership
without realzing it since that is not the "normal" behavior of
operator=. I'll have to research that more before I implement it, and I
had thought about implementing it also but am not aware of all the
effects. Disabling operator= and creating the move constructor allow me
to use the object as intended. Only one real copy of the object
exists, almost like a singleton pattern.

Bo Persson

You are indeed mistaken. One of the reason rvalue refs are in the upcoming
C++0x standard is because they make moving semantics safe. Consider
auto_ptr, where ownership *can* accidentally be lost. On the other hand,
this can't happen with unique_ptr (its replacement). To move a unique_ptr
through assignment from an lvalue, you have to request it, as in:

std::unique_ptr<T> original = /* initialize here */
std::unique_ptr<T> dest = original; // won't compile
std::unique_ptr<T> safe_dest = std::move(original);
// don't use original anymore

So if you class has an operator=(&&), but not a copy assignment operator
(operator=(const&), but also operator=(&)), you can't accidentally
invalidate your objects. Although check that your implementation actually
*does* that since the specs for rvalue refs have been in flux for some

I just tried it out and it seems that in Express 2010 anyway my concern was
justified. Without an operator=(&&) this will not compile (as it shouldn't).

world.fonts[L"Normal"] = jmlGL::jglFont( hDC, L"Courier New", 24 );
error C2248: 'jmlGL::jglFont::operator =' : cannot access private member
declared in class 'jmlGL::jglFont'

However, when I add
jglFont& operator=(jglFont&& rhs) { /*.real code here..*/ }

my program compiles and runs. I did not have to use std::move.

As a final note, don't forget that std::move is an innocent call: in my
example, it resolves to a static_cast<T&&>. It doesn't actually do any
move: it's up to your operator=(&&) to do the right thing (i.e. it can
safely assume that it can pilfer its argument), but the compiler won't do
anything for you (unless you default the operator if this is supported).

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