Re: g++ weird behavoir with values returned by copy.

[Nathan Sidwell <nathan at codesourcery dot com>]

Leandro Lucarella wrote:
>

> Is tmp who is returned, not a copy. This works fine on most cases, but
> I've found a case where the behavoir is much more weird (at least for
> me).
>
> and I found a very weird case, in wich with cause an object to be
> destroyed when it's still in use using a reference:
Not a bug. This is C++. The optimization you are observing is the
Return Value Optimization, and is not directly connected to the
effect you are seeing.


> struct A {
>         A(void) {}
>         A(const A& a) {}
>         ~A(void) {}
>         A& operator=(const A& a) {
>                 return *this;
>         }
>         A& print(void) {
>                 return *this;
>         }
> };
>
> A test(void) {
>         A tmp;
>         cout << "Doing something." << endl;
>         return tmp;
> }
>
> int main(void) {
>         A& ar = test().print();
> 	// Now ar points to a destroyed objec
yes, as the std specified.
	test () returns a temporary object, which is used in the call to A::print.
that returns a reference to self, which is bound to ar. As ar is not directly bound
to the temporary A that test returned, that is destroyed at the end of that full
expression.

you need to write
	A &ar = test ();
	ar.print ();

now, ar is directly bound to a temporary, and that temporary's lifetime is
extended to be the same as the reference to which it is bound.

nathan
--
Nathan Sidwell    ::   http://www.codesourcery.com   ::     CodeSourcery LLC
         The voices in my head said this was stupid too
nathan@codesourcery.com    ::     http://www.planetfall.pwp.blueyonder.co.uk