Re: Mutability of temporary variables

From:

"Paul" <pchrist<nospam>or@yahoo.co.uk>

Newsgroups:

comp.lang.c++

Date:

Sun, 20 Nov 2011 00:23:22 -0000

Message-ID:

<ja9h9v$t2t$1@adenine.netfront.net>

"Victor Bazarov" <v.bazarov@comcast.invalid> wrote in message
news:ja94s6$61c$1@dont-email.me...

On 11/19/2011 9:20 AM, Paavo Helde wrote:

Victor Bazarov<v.bazarov@comcast.invalid> wrote in
news:ja8bo7$88k$1@dont-email.me:

On 11/19/2011 7:33 AM, kyle wrote:

Consider following code:

int main() {
const int& c = int();
int& m = const_cast<int&>(c);
m = 4;
}

The object of the snippet is to get a mutable reference to a
temporary. This cant be done directly because non-const reference
cannot bind to temporary, but we should be OK with casting away
constness of reference 'c' since it doesn't actually refer to const
object.

Temporaries are mutable, so in whole my snippet is legal C++. Am i
correct?

AIUI, your code has undefined behavior. 'c' is bound to an rvalue,
and using 'const_cast' to produce an lvalue out of it is not among the
allowed operations for 'const_cast'.

The lvalue and rvalue notions apply to source code expressions (3.10/1:
"Every expression is either an lvalue or an rvalue"). Expression 'c' is
of a reference type, which means it is a lvalue, doesn't it? (rvalue
references use&& syntax). And lvalue->lvalue conversion is well defined
for const_cast.

Or have I misunderstood something?

int() is an rvalue expression. Now, explain that.

What is it about "int()" that you do not understand exactly?

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