Re: Mutability of temporary variables
Victor Bazarov <v.bazarov@comcast.invalid> wrote in 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.
So is A() (where A is a class type for example), so by that logic it
would not be possible to create mutable temporaries at all.
In the OP-s example the const_cast was not applied to int(). This would
be indeed an error (Comeau agrees:)
"ComeauTest.c", line 2: error: expression must be an lvalue
int& m = const_cast<int&>(int());
const_cast was applied to a reference instead. Reference always points to
an object; yes, this may be some temporary object initialized by an
rvalue, but so what? Casting away constness would be illegal if the
object is const (as suggested by Johannes), but this is another topic and
not related to rvalues at all.
Cheers
Paavo