Re: why is binding non-const ref to return value bad?

From:
peter koch <peter.koch.larsen@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 10 Mar 2010 13:48:56 -0800 (PST)
Message-ID:
<c65b4400-43e0-446f-a095-37dee32e79e1@g28g2000yqh.googlegroups.com>
On 10 Mar., 18:08, PeteUK <newbar...@gmail.com> wrote:

Hello,

I'm doing a code review for someone and picked them up on this:

BigClass AFunction() { /*returns a large object by value*/ }

Caller()
{
  BigClass bc = AFunction();
  /* use bc */

}

I saif they should do this in the caller to avoid copying the
temporary:

Caller()
{
  const BigClass& bc = AFunction();
  /* use bc */

}

They came back with the following code which compiled but I asked them
to change to const but I had trouble justifying why it should be
const:

Caller()
{
  BigClass& bc = AFunction();
  /* use bc */

}

Does the life of the temporary get extended using the non-const
reference as it does from using the const reference? Was I right to
pick him up on it not being const? Please give me some rationale if
I'm right!


Well, first of all I believe you should know C++ better to do a code
review. First of all, you misunderstand the consequence of extending
the life of the temporary. By having a const& you essentially have the
code

 BigClass __hidden = AFunction();
 BigClass const& bc = __hidden;

Secondly you seem to be unaware of RVO. There will be no copy of the
big object in the code shown - at least not with any compiler I know
of when you optimise the code.

Last, you are complaining about some code because of some undefined
performance concern. To do this without doing measurements is quite
na=EFve.

/Peter

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