[boost] Re: Re: [test_tools] Problem with struct wrapstrstream.

From: Anthony Williams (anthwil_at_[hidden])
Date: 2002-10-01 03:46:12

Gennadiy Rozental writes:
 > > Difficult but I'll try. Rvalues of non-class types don't have addresses as
 > > they can live outside the addressable memory (in registers, for example.)
 > > When a reference is bound to an rvalue, that rvalue may need to be copied
 > to
 > > addressable memory. In a context where 'this' is not used, rvalues of
 > class
 > > types can live outside addressable memory, too.
 > 
 > Does this mean that if 'this' IS used than compiler is not allowed to put
 > rvalues of class types outside addressable memory?

 You can call member functions on rvalues of class type. If the member
 function uses this, then it must be addressable.
 
 > > So the compiler in the above is allowed to use the copy constructor to
 > > create another (addressable) temporary of type 'A' (or 'A const') and bind
 > > the reference to it, instead of binding directly to 'A()' that may not be
 > > addressable.
 > Continue my question: does the following should compile?
 > 
 > class A {
 > public:
 >     A() {};
 >     int field;
 > private:
 >     A( A const& );
 >     void operator=( A const& );
 > };
 > 
 > void foo( A const& a )
 > {
 >     int i = a.field;
 > }
 > 
 > void moo()
 > {
 >     foo( A() );
 > }

No. The copy constructor must be accessible, so the temporary A() can be bound
to the const reference. Note that in doing such a binding, the compiler may
make an _arbitrary_ number of copies (e.g. 10000000) if it feels like it, or
it may be able to omit all copies. To cater for the first case, the copy
constructor _must_ be accessible. If the compiler doesn't actually make a
copy, it doesn't have to be defined though.

Anthony