Re: Mutability of temporary variables

From:

Leigh Johnston <leigh@i42.co.uk>

Newsgroups:

comp.lang.c++

Date:

Sat, 19 Nov 2011 14:25:30 +0000

Message-ID:

<Do2dnVH-ff_DIVrTnZ2dnUVZ8qOdnZ2d@giganews.com>

On 19/11/2011 13:45, Victor Bazarov wrote:

On 11/19/2011 8:18 AM, Paavo Helde wrote:

kyle<kyle@nomail.com> wrote in
news:op.v4605gilpbcp0d@localhost.localdomain:

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?

Yes, I think you are correct. Another trick to get a mutable reference to
a temporary is to use a non-const member function. This avoids
const_cast, but of course lifetime extending by binding to a const
reference does not work any more:

struct A {
int m;
A& Ref() {return *this;}
};

void f(A& a) {
a.m = 4;
}

int main() {
f( A().Ref() );
}

There is a difference between your example and the OP's. In your example
the temporary is of a class type, and the expression A() produces an
lvalue to begin with.

No; A() produces an rvalue.

/Leigh