Re: MSDN const_cast sample

From:
Abhishek Padmanabh <abhishek.padmanabh@gmail.com>
Newsgroups:
microsoft.public.vc.language
Date:
Tue, 18 Dec 2007 09:56:23 -0800 (PST)
Message-ID:
<a47eb267-db06-47ef-b21a-59c9ef9fbae3@s8g2000prg.googlegroups.com>
On Dec 18, 7:56 pm, "Ben Voigt [C++ MVP]" <r...@nospam.nospam> wrote:

"Abhishek Padmanabh" <abhishek.padman...@gmail.com> wrote in message
news:3daa8feb-f9ca-4572-8815-72933b7704bc@e6g2000prf.googlegroups.com...

On Dec 18, 12:16 am, "Ben Voigt [C++ MVP]" <r...@nospam.nospam> wrote:

Otherwise, a compiler is non-compliant if casting away const doesn't
work.
Assuming this code:

  void mutate( int const& i) {
    const_cast<int&>(i) = 42;
  }

  int i = 5;
  mutate(i);
  std::cout << i << std::endl;

I would assume the compiler to behave properly and output 42. Do you
know
of
any compilers that don't?


It should. It is only undefined behaviour if the variable/object being
modified was originally created as const. Here, i is not. And hence,
the modification should work. Don't know of a compiler that would go
against the above though.


I don't think "originally created as non-const" is sufficient.

For example:

void mutate( int const& i) {
   const_cast<int&>(i) = 42;

}

void f(const int* const p)
{
    int j = *p;
    mutate(*p);
    std::cout << *p << j << std::endl;

}

int i = 5;
f(&i);

I see no reason to believe that the compiler won't use the register used
to
hold j instead of reading memory again. For that matter, it is probably
allowable to delay assigning a value to j until after the call to mutate.


No reason until there is no const_cast applied. Herb Sutter's


I don't understand that statement.

All I know is that f isn't aware of the internal working of mutate, it just
knows the contract, which is that mutate receives a reference to const.
Therefore f can assume that *p cannot change and therefore can cache its
value throughout the function.


Yes, that is the contract and it is true that a function that takes a
reference to const and casts away const and modifies it is bad. But
the point I am making is that if the object when declared as non-
const, is passed to such a function, it is not undefined behavior. It
is only undefined behaviour if the object passed to this function was
a const object (const declaration for creating it). I am not saying
such functions are a good programming style, its bad bad bad since f()
can not really know if the constified object is actually a const or
not (may be with some effort but it won't make it a good function).

Here is a reference - http://groups.google.co.in/group/comp.lang.c++.moderated/msg/955f2471ef8d7801
- Note that the call f(poly) is fine. It does not result in undefined
behaviour because poly is not a const object. But cpoly is and hence
f(cpoly) is undefined behaviour.

I guess, when in f(), as soon as a const_cast is applied and the
object is non-const, the compiler has to make sure that it writes the
change at the memory where the object resides and not the cached value.

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