Re: Trying to understand "&&"

From:
Chameleon <cham_gss@hotmail.com>
Newsgroups:
comp.lang.c++
Date:
Sun, 02 Jan 2011 14:59:46 +0200
Message-ID:
<ifpss5$oeu$1@news.grnet.gr>
???????? 01/01/2011 22:02, ??/?? SG ????????????:

On 1 Jan., 14:42, Chameleon<cham_...@hotmail.com> wrote:

class A
{
public:
     A();
     A(const char *string);
     ~A();
     A(A&a);


The copy constructor usually takes a reference to a CONST.


Yes, Indeed.

     A&& operator+(A&a) {
         A b;
         b.str = new char[strlen(this->str) + strlen(a.str) + 1];


This this->str as well as a.str could be null pointers couldn't they?


Of course. But it is a very very super fast sample. So I don't check
every case.

         strcpy(b.str, this->str);
         strcpy(b.str + strlen(this->str), a.str);
         cout<< "A + A = "<< b.str<< endl;
         return std::move(b);
     }


A&& is a reference type. b is a function-local object. You're
returning a reference to a function-local object. This produces a
dangling reference. Using this dangling reference invokes undefined
behaviour. When in doubt, avoid declaring functions that return rvalue
references. This is almost always the wrong thing to do. Your operator
+ should "return an object by value". Also, if you return a function-
local object you should not use std::move as this inhibits a potential
copy/move elision.

Cheers!
SG


Thanks a lot!
After 8 years of C++ programming, I realize that return an object by
value, doen't invoke the D'tor of the function-local object.
Whow!!!
This is what I am trying to do, but it exists!.

So, the "&&" needed in case where you must put an object in a
vector<UberObjectClass>. Until now, I used vector<UberObjectClass*>.
Am I right??

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