Re: Temporary objects and operators overloading

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 9 Jul 2009 13:17:48 CST
Message-ID:
<dc24c44b-1933-4aca-8f44-bf414c7d86d2@d4g2000yqa.googlegroups.com>
Francis Glassborow wrote:

James Hopkin wrote:

Francis Glassborow wrote:

SG wrote:

I actually tested it a couple of weeks ago with G++ 4.3.3:
  Foo operator+(Foo const & lhs, Foo const & rhs){
     return Foo(lhs) += rhs; // No RVO is done
  }

You mean it called the copy ctor twice? I am surprised.


Looking at 12.8/15 (the RVO):

  when a temporary class object that has not been bound to a
  reference (12.2) would be copied to a class object with the
  same cv-unqualified type, the copy operation can be omitted
  by constructing the temporary object directly into the
  target of the omitted copy

it explicitly rules out eliding the copy when there's a reference
involved. The compiler *has* to call the copy constructor twice
(modulo the 'as if' rule).


But Foo(lhs) += rhs;
makes no attempt to bind the temporary being used to a reference.


Does "return *this;" in operator+= count as creating a reference?

Anyhow, I think I explained it well from a practical point of view.
There is no difference between the following two:

  return Foo(lhs) += rhs;
  return Foo(lhs).operator+=(rhs);

where operator+= is a function that returns a reference.
Unfortunately, the function signature

  Foo& Foo::operator+=(Foo const&);

doesn't promise that the function returns *this. Therefore, the
compiler doesn't know in advance to what object the return value will
refer to. Because of that, it has to copy construct the return value
from whatever operator+= returns and allocate additional space for the
temporary object Foo(lhs).

In case operator+= is inlined the compiler could do better. Whether
that elision is legal is a question I cannot answer.

Cheers!
SG

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