Re: MSDN const_cast sample

From:
"Ben Voigt [C++ MVP]" <rbv@nospam.nospam>
Newsgroups:
microsoft.public.vc.language
Date:
Mon, 17 Dec 2007 13:16:57 -0600
Message-ID:
<O7Sy4EOQIHA.1188@TK2MSFTNGP04.phx.gbl>

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.

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