Re: Are literals objects?

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 6 Apr 2007 14:23:23 -0400

Message-ID:

<ev636s$e9j$1@news.datemas.de>

Greg Comeau wrote:

In article <ev5pgd$lee$1@news.datemas.de>,
Victor Bazarov <v.Abazarov@comAcast.net> wrote:

SasQ wrote:

[..]
Literals are r-values and some expressions [yielding temporary
objects] are r-values too, so it's impossible to get their
addresses.

We're going off on a tangent here, but objects of class types have
addresses even if they are r-values. Behold:

   #include <iostream>
   struct foo {
       void bar() { std::cout << this << std::endl; }
   };
   int main() {
       std::cout << "Temporary at " << foo().bar();
   }

Isn't this trying to output the value of the pointer to member?

No, but I did type it too quickly and didn't test. It ought to be

    int main() {
         std::cout << "Temporary at "; foo().bar();
    }

Which I don't think is defined (IOWs, the cout in main doesn't
compile?)?

Right.

V
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