Re: MSDN const_cast sample
"George" <George@discussions.microsoft.com> wrote in message
news:DCC0E52E-4BBC-46F8-BEC0-6B552C760335@microsoft.com...
Hi Ben,
I have used Visual Studio 2008 tested your code works. The output is 425.
Under debug or release? Which optimizations enabled, if release?
I do not quite understand what do you mean,
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.
What do you mean *won't use the register* and *delay assigning*? Is it
some
special approaches to do optimization? Could you provide more descriptions
please?
regards,
George
"Ben Voigt [C++ MVP]" 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.
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