http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50056
--- Comment #2 from Wolfgang Roehrl <wolfgang.roe...@gi-de.com> 2011-08-12
14:51:25 UTC ---
Hi Jonathan,
I would analyze the crucial line of the code as follows:
The cast creates a temporary reference, which binds to the temporary
object,
and then initalizes 's' with the temporary reference. The latter means:
A) 's' is initialized with a temporary S-object which appears as an
lvalue:
Cf. 5/6: "If an expression initially has the type "reference to T"
(8.3.2,
8.5.3), the type is adjusted to "T" prior to any further analysis, the
expression designates the object or function denoted by the reference,
and the expression is an lvalue."
B) 's' is directly bound to the temporary S-object:
Cf. 8.5.3/5, Bullet 1: "If the initializer expression
- is an lvalue (but is not a bit-field), and "cv1 T1" is reference-
compatible with "cv2 T2," ... then the reference is bound directly
to the initializer expression lvalue."
C) 12.2/5 says that the lifetime of a temporary which is bound to a
reference
is extended to the lifetime fo the reference.
I'm not sure if this analysis is really correct but it seems reasonable to
me.
Otherwise we would have a reference without an object.
Best regards,
W. Roehrl
"redi at gcc dot gnu.org" <gcc-bugzi...@gcc.gnu.org>
12.08.2011 14:47
An
wolfgang.roe...@gi-de.com
Kopie
Thema
[Bug c++/50056] Binding a temporary object to a reference
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50056
--- Comment #1 from Jonathan Wakely <redi at gcc dot gnu.org> 2011-08-12
12:46:42 UTC ---
Are you sure G++ isn't correct?
The cast creates a temporary reference, which binds to the temporary
object,
then initalizes 's' with the temporary reference. That means the temporary
object's lifetime is extended to the same lifetime as the reference, but
the
reference is temporary so is destroyed at the end of the full expression.
call S::S()
bind to temporary reference
initialize s from temporary reference
temporary reference goes out of scope
call S::~S()
call S::func()
s goes out of scope
if you do this:
const S& s = S();
then you get the behaviour I assume you are expecting:
call S::S()
initialize s temporary object
call S::func()
s goes out of scope
call S::~S()
