Re: Get lvalue through rvalue

From:

"Igor Tandetnik" <itandetnik@mvps.org>

Newsgroups:

microsoft.public.vc.language

Date:

Mon, 17 Dec 2007 09:14:49 -0500

Message-ID:

<egRWwcLQIHA.5160@TK2MSFTNGP05.phx.gbl>

"George" <George@discussions.microsoft.com> wrote in message
news:8FAE36E4-5C6F-4779-9E04-955E34074AF7@microsoft.com

I do not know how in the following code, rvalue -- return of X(),
could result in a lvalue finally and binded to a non-const reference
input parameter of function f.

Any ideas?

[Code]
struct X {

};

void f (X& x) {}

int main()
{
f (X() = X());

return 0;
}
[/Code]

There's an implicitly declared compiler-generated assignment operator in
X, that looks like this:

struct X {
X& operator=(const X&) { return *this; }
};

So f(X() = X()) is really

f( X().operator=( X() ) );

Now, it is legal to call non-const member functions on temporaries, and
a temporary can be passed as a const X& parameter. Since operator=
returns a non-const reference, the whole thing works similarly to
lvalue_cast we've discussed in previous threads, only less efficiently.
--
With best wishes,
Igor Tandetnik

With sufficient thrust, pigs fly just fine. However, this is not
necessarily a good idea. It is hard to be sure where they are going to
land, and it could be dangerous sitting under them as they fly
overhead. -- RFC 1925