Re: Temporary Object Lifetime

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 13 Jun 2008 02:47:49 -0700 (PDT)
Message-ID:
<5adfff6e-7d41-4005-98cb-e47f3ee6ce06@26g2000hsk.googlegroups.com>
On Jun 12, 10:40 pm, Victor Bazarov <v.Abaza...@comAcast.net> wrote:

better_cs_...@yahoo.com wrote:

class Foo {/* Details don't matter */};

class Bar {
public:
   Bar(): m_Foo(/* Construct a Foo however it wants to be constructed
*/);
   const Foo &GetFoo() const { return m_Foo; }
private:
   Foo m_Foo;
};

int main() {
   const Foo &OneFoo = Bar().GetFoo();
   // Do stuff with OneFoo. Is this legal given the use of
   // temporaries above?
}


I don't believe so. The 'Bar' temporary is not bound to the
reference, so it's not going to live beyond 'OneFoo's
initialisation.


That's irrelevant. The fact that a reference is bound to a
temporary has no effect on the lifetime of the temporary. The
only time the lifetime of the temporary is extended is when the
temporary is used to initialize the reference, and that's not
the case here: the reference in main is initialized with another
reference, and the reference returned by Bar::GetFoo() is
initialized by an lvalue, not a temporary (and reference return
values are a special case anyway, and don't extend the lifetime
of the temporary, even if they're initialized with one).

Had 'Bar' derived from 'Foo' *and* you did

     const Foo &OneFoo = Bar();

, then the 'Bar' would have a subobject of type 'Foo' that ought to
survive longer than the full expression, and as the superobject to the
temporary bound to a reference, the 'Bar' temporary would have to stay
alive as long as 'OneFoo'.

But your case is not that. In your case you create a
reference to a data member (which does live even shorter than
the 'Bar' object that contains it), so as soon as the
initialisation is complete, the 'OneFoo' reference becomes
*invalid*.


The key isn't whether something is a subobject or not. In his
initial example, the reference is bound to a sub-object of the
temporary as well. The key here is that the temporary is
initializing the reference directly; if m_Foo were public and
you wrote:

    Foo const& &OneFoo = Bar().m_Foo ;

it would be fine as well. It's the indirections in his case
which break the code.

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James Kanze (GABI Software) email:james.kanze@gmail.com
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His body a blood-ruby radiant
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