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From: "wolfgang.roehrl@gi-de.com" <gcc-bugzilla@gcc.gnu.org> To: gcc-bugs@gcc.gnu.org Subject: [Bug c++/50056] Binding a temporary object to a reference Date: Fri, 12 Aug 2011 15:07:00 -0000 [thread overview] Message-ID: <bug-50056-4-uia7PLHE70@http.gcc.gnu.org/bugzilla/> (raw) In-Reply-To: <bug-50056-4@http.gcc.gnu.org/bugzilla/> http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50056 --- Comment #2 from Wolfgang Roehrl <wolfgang.roehrl@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-bugzilla@gcc.gnu.org> 12.08.2011 14:47 An wolfgang.roehrl@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()
next prev parent reply other threads:[~2011-08-12 14:51 UTC|newest] Thread overview: 4+ messages / expand[flat|nested] mbox.gz Atom feed top 2011-08-12 12:39 [Bug c++/50056] New: " wolfgang.roehrl@gi-de.com 2011-08-12 13:27 ` [Bug c++/50056] " redi at gcc dot gnu.org 2011-08-12 15:07 ` wolfgang.roehrl@gi-de.com [this message] 2011-08-16 23:55 ` jason at gcc dot gnu.org
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