On Apr 4, 1:22 am, Ron Natalie <r...@spamcop.net> wrote:
> Lilith wrote:
> > Does the standard define the order in which objects declared
> > in the same scope are destructed when those objects go out
> > of scope? This has a bearing on some cleanup work I want to
> > incorporate in one of my classes.
> Scope has no meaning for object lifetime. It doesn't exist at
> runtime.
Yes and no. The concepts are definitely orthogonal, but for
historical reasons, in C++, object lifetime, or at least the
default object lifetime, is partially determined by the scope of
the object's definition.
> But, with the exception of the dynamic allocation where you
> specifically have control over when you destroy it, objects
> are always destructed in reverse order of construction.
Only within very restricted categories, and with numerous
exceptions. Reverse order is true for members of a class,
temporaries whose lifetime hasn't been extended, non-static
local variables, objects with static lifetime, and I think (but
I'm not sure) thrown objects. It's trivial, however, to create
examples where order of destruction is not the reverse of
construction: anytime a temporary is bound to a const reference,
for example, or is used to initialize an other object; or between
static and non-static local variables. E.g.:
void
f()
{
MyClass a ;
static MyClass b ;
}
Order of construction: a, then b. Order of destruction, a, and
sometime well after leaving f, b. (Throw in a call to exit() at
the end of f, and the issue becomes even less clear---a will
never be destructed.)
--
James Kanze (GABI Software) email:
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