Order of Object destruction

[Eljay Love-Jensen]

Hi Neophytos,

Yes, the order of destruction is specified, due in no small part to the 
lifespan of a temporary being well-defined.

Effectively, the temporary object lives until the end-of-statement semicolon.

As I understand it (I've never needed to tried this), if there were an 
alias to it, it would live until the end of the extant of alias.  For example:
A a;
A& b = foo(a);
// a and b are live.
cout << a.y << endl;
cout << b.y << endl; // Okay.

Before the standard's committee nailed the issue, compiler vendors were 
inconsistent when the temporary object was destructed.  Which lead to all 
sorts of non-portable code and frustrated programmers.

--Eljay