Re: Passing References: Is this correct and according to standard?

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Sun, 13 Jul 2008 19:57:39 +0200
Message-ID:
<-I-dnVVDnoSL3-fVnZ2dnUVZ_rTinZ2d@posted.comnet>
* James Kanze:

On Jul 11, 1:10 am, "Alf P. Steinbach" <al...@start.no> wrote:

* Akshay Loke:

I have this function from a class MFnDagNode,
   addChild( MObject & child, unsigned int index = kNextPos, bool
keepExistingParents = false );
which takes an object reference as first input parameter (ignore the
rest since those are default)

now in another class, I have:

MFnDagNode parent;
parent.addChild(node.object());
   Where node is MfnDependencyNode&
   And node.object() returns MObject

This gives me error
?error: no matching function for call to
?MfnDagNode::addChild(Mobject)?
   Note: candidates are: Mstatus MFnDagNode::addChild(MObject&, unsigned
int, bool)

Now I am not sure why this doesn?t work and the object
doesn?t get implicitly cast to the MObject& reference
argument, but I am assuming that could be because the
node.object() returns a temporary const object and it cannot
be cast to a MObject& reference?


Probably it doesn't return a temporary const object. Probably
it returns an object. Which is not const but is an rvalue.

So this is the way I have done this?

const MObject& nodeObjectConstRef = node.object();
MObject& nodeObjectRef = const_cast<MObject&> (nodeObjectConstRef);
parent.addChild(nodeObjectRef);

and This works!


Assuming node.object() returns by value, it's formally UB.


Unless the return type is declared const, I don't see the
undefined behavior. Could you explain?


An originally const object can't be modified without formal UB. In the case of
return by value the declaration of nodeObjectConstRef declares a reference to an
originally const (temporary) object.

Why don't you do

   MObject nodeObject = node.object();
   parent.addChild( nodeObject );

?


More to the point, why doesn't he declare the function to take a
reference to const?


I don't think the code really makes sense.

Cheers,

- Alf

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