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returning a (const) reference to a temporary

 
 
James Kanze
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      02-22-2011
On Feb 22, 9:21 am, AdlerSam <(E-Mail Removed)> wrote:
> On 22 Feb., 10:07, Paul Brettschneider <(E-Mail Removed)>
> wrote:
> > AdlerSam wrote:


> > > I wonder why the following two lines produce a warning:


> > > class X {};
> > > const X &f() {return X();}


> > > $ g++ -c ref.cpp
> > > ref.cpp: In function ‘const X& f()’:
> > > ref.cpp:2: warning: returning reference to temporary


> > > As far as I understand, a const reference _extends_ the
> > > lifetime of a temporary until the very last reference
> > > instance that refers to the temporary goes out of scope.


That would require full garbage collection, and then some, in
order to implement.

> > > Thus, where is the problem that justyfies the warning?


> > This assumption is - of course - nonsense. If you want to
> > manage lifetime of objects, you usually use smart pointers
> > or containers.


> > Hope that helps.


> Hm - Then where do I have mistaken Herb Sutters GotW #88


> http://herbsutter.com/2008/01/01/got...r-the-most-imp...


> To quote the important part:


> > Normally, a temporary object lasts only until the end of the
> > full expression in which it appears. However, C++
> > deliberately specifies that binding a temporary object to
> > a reference to const on the stack lengthens the lifetime of
> > the temporary to the lifetime of the reference itself, and
> > thus avoids what would otherwise be a common dangling
> > reference error. In the example above, the temporary
> > returned by f() lives until the closing curly brace. (Note
> > this only applies to stack-based references. It doesn’t work
> > for references that are members of objects.)


You seem to misunderstand two important points:

-- first, the lifetime of the temporary is extended only to the
end of the lifetime of the reference the temporary
initializes, not to any other references, and

-- second, return involves a copy, so the reference being
returned is not the reference the temporary initialized.

Herb has simply used a common, but misleading formulation of the
rule, which is better stated as "initializing a reference with
a temporary extends the lifetime of the temporary to that of the
reference". When returning a reference, you (formally, at
least) initialize a local temporary reference with the return
expression, then return a copy of that reference, with the local
temporary reference going out of scope (and thus triggering the
destruction of the temporary used to initialize it).

--
James Kanze
 
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