Re: C++0x: returning rvalue references, recycling temporaries

From:
Seungbeom Kim <musiphil@bawi.org>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 13 Sep 2008 21:09:04 CST
Message-ID:
<gafsao$hph$1@news.stanford.edu>
Howard Hinnant wrote:

Now consider:

string operator+(const string&, const string&);
string operator+(string&& x, const string& y) {x += y; return
move(x);}

const string& d = a + b + c;

This is slightly more expensive because it involves an extra move
construction. However "a + b + c" now creates a temporary whose
lifetime is going to extend beyond the semicolon until the reference d
is destructed due to the language in 12.2 [class.temporary].

For the cost of an extra move construction (after an append) you can
sleep better at night. :-)


1. How is it more expensive by involving an extra move construction?
As far as I understand, move(x) above is just a shorthand for
static_cast<string&&>(x). Does the static_cast incur any (significant)
operation in the machine code level?

2. Why do we need the move? It converts an lvalue reference to an rvalue
reference, but we are returning by (r)value anyway, and we can always
safely return a reference when the return type is not a reference and
that creates a copy of the referent; is this correct?

     T foo(T& t) { return t; } // returns a *copy* of t
     T foo(T&& t) { return t; } // what about this?

What about not writing the explicit move: is it correct/desirable?

     string operator+(string&& x, const string& y) { x += y; return x; }

--
Seungbeom Kim

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