RE: [Boost-Users] Re: "Correct" way to return empty shared_ptr?

From: dick.bridges_at_[hidden]
Date: 2003-04-24 11:50:00

This is one of the places where I get *REALLY* confused - the 'real' code
has multiple returns.  Apologies:  I should have included this psuedo code
in the original example.  It looks like this.

<snippet>
// v is populated somewhere else
typedef vector<shared_ptr<int> > V;
V v;

shared_ptr<int> maybe_get_a_pointer()
{
    shared_ptr<int>::iterator itr;
    itr = find_if(v.begin(), v.end(), some_op);

    if(itr != v.end)
        return(*itr);

    shared_ptr<int> p;
    return(p);
}
</snippet>

Are compilers supposed to be able to match *itr to p in order to apply the
NRVO?

                                                                                                                                                  
                      "Ben Hutchings"                                                                                                             
                      <ben.hutchings_at_businesswebso         To:      <Boost-Users_at_[hidden]>                                                 
                      ftware.com>                          cc:                                                                                    
                                                           Subject: RE: [Boost-Users] Re: "Correct" way to return empty shared_ptr?               
                      04/24/2003 08:56 AM                                                                                                         
                      Please respond to                                                                                                           
                      Boost-Users                                                                                                                 
                                                                                                                                                  
                                                                                                                                                  

dick.bridges_at_[hidden] wrote:
> I don't understand C++ well enough to grok a compiler's interpretation
of
> the code, but I seems like I'm creating an extra copy for the return.
<snip>

Logically the local variable is copied to the the caller, but since the
type
of the returned expression matches the return type the compiler is
allowed
to optimise away that copy, even though the return type has a
user-defined
copy constructor.  In this case you always return a certain named
variable,
so the compiler can arrange to construct it in the memory location where
the
caller expects the returned value.  This is called the
named-return-value
optimisation, or NRVO.

Info: <http://www.boost.org>
Wiki: <http://www.crystalclearsoftware.com/cgi-bin/boost_wiki/wiki.pl>
Unsubscribe: <mailto:boost-users-unsubscribe_at_[hidden]>

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/