From -7273212585619446453 X-Google-Thread: f78e5,ff39f62da1e60460,start X-Google-Attributes: gidf78e5,public X-Google-Language: ENGLISH,ASCII-7-bit Path: g2news1.google.com!news4.google.com!news2.google.com!proxad.net!news.cs.univ-paris8.fr!newsfeed.vmunix.org!feed.news.tiscali.de!newsfeed.icl.net!newsfeed.fjserv.net!peer-uk.news.demon.net!kibo.news.demon.net!news.demon.co.uk!demon!stump.algebra.com!devnull From: stkim@yujinrobot.com ("Kim, Seungtai") Newsgroups: comp.std.c++ Subject: Temporary object's lifetime and Full-expression Date: Fri, 22 Apr 2005 19:44:48 GMT Organization: Korea Research Environment Open Network Lines: 108 Sender: mail2news@demon.net Approved: fjh@cs.mu.oz.au (Fergus Henderson , moderator of comp.std.c++) Message-ID: NNTP-Posting-Host: news.news.demon.net X-Trace: news.demon.co.uk 1114199096 17121 158.152.254.254 (22 Apr 2005 19:44:56 GMT) X-Complaints-To: abuse@demon.net NNTP-Posting-Date: Fri, 22 Apr 2005 19:44:56 +0000 (UTC) X-Robomod: STUMP, ichudov@algebra.com (Igor Chudov) X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2600.0000 X-Priority: 3 X-Virus-Scanned: by amavisd-new at cs.mu.OZ.AU X-MSMail-Priority: Normal X-Received: (from fjh@localhost) by mulga.cs.mu.OZ.AU (8.12.10+Sun/8.12.9/Submit) id j3MJimlL004316; Sat, 23 Apr 2005 05:44:48 +1000 (EST) X-Path: comp-std-cpp-robomod!not-for-mail X-NNTP-Posting-Date: Thu, 21 Apr 2005 08:38:56 +0000 (UTC) X-Delivered-To: std-c++@ucar.edu X-Authentication-Warning: mulga.cs.mu.OZ.AU: fjh set sender to devnull@stump.algebra.com using -f X-Newsreader: Microsoft Outlook Express 6.00.2600.0000 X-Newsgroups: comp.std.c++ Xref: g2news1.google.com comp.std.c++:4565 There were some discussion about parameter passing of the temporary object at the han.comp.lang.c++. The original question was simple why bellow code dose not work. foo( &A() ); Yes. Clear. It's illegal use of operator & that requires the lvalue which the tempporary dose not return. And there another question was followed. "What about this case? Is it legel and dose it produce well-defined behavior?" struct A { A& me() { return *this; } }; void foo(const A* a); ... foo( &A().me() ); I said it's not. It will make the undefined behavior. The reason is that the temporary object will be destroyed at the point of immediately after calling function foo. Therefore foo will get the dangling pointer. It's illegal use of invalid pointer. But the others assist the opposite opinion that the temporary object will not be destroyed until the completion of the full expression calling foo. My opinion comes from the definition of the temporary object's lifetime and the full-expression. 12.2/p3 states the general rule for lifetime of the temporary object. 12.2/p3 Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. And 1.9/p12 states the definition of the full-expression. 1.9/p12 A full-expression is an expression that is not a subexpression of another expression. And 1.9/p13 describes the function call is full expression, also. 1.9/p13 ... the function call is a full-expression. In this point, there can be two different but ciritical interpretations whether the function call in an expression is a sub-expression or not it's an independent full expression. - The function call is sub-expression. It's explicitly included in the other expression. So, the temporary object should not be destroyed until after evaluating full expression that contains the function call. Therefore OP's above code is legal and produce well-defined behavior. - The function call is full expression that it is independent from the expression containing the function call. For that reason, the temporary object shold be destroyed immediately after calling function. Hance, above code will make UB. Which opinion is correct? I have the more questions if we assume that the function call is a sub-expression of the other expression. 1. Why dose the bellow sentance exist? 12.4/p5 ... A temporary bound to a reference parameter in a function call(5.2.2) persists until the completion of the full expression containing the call. ... If function call is a sub-expression, it's not required. Always the temporary object survives until the completion of the full expression containing the function call. Why it needed? Am I something wrong and mis-interpretation? 2. Why dose the 1.9/p13 emphasize "function call is a full expression." twice. If we think about that there are return of the function, it's clear that the function call alone could not be the full-expression, always. For example, in a "a = f( x ) ;", f( x ) is just a sub-expression, not a full-expression. Another case, even f returns void and in expression of "f( x );", the final expression is "void", not a function call. For that reason, a function call itself can not be the full-expression. And the explanation should be just a "expression" not a "full expression". Is it overlooking? 3. Last one. Is there full expression in the C++ program if the function call is just a sub-expression? I think there isn't. If we define the function call makes the relation of the sub-expression, all the reacheable functions comes from the main and creation realated functions for the static obejcts. And they are called by execution environment. So, there is no full-expression. Is it true? Thanks in advice. -- S Kim --- [ comp.std.c++ is moderated. 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