Re: Overloaded vendor library routine: is this C++? Or very clever C?

From:

"Noah Roberts" <roberts.noah@gmail.com>

Newsgroups:

comp.lang.c,comp.lang.c++

Date:

15 May 2006 14:48:50 -0700

Message-ID:

<1147729730.251440.106550@j55g2000cwa.googlegroups.com>

Keith Thompson wrote:

[Followups redirected to comp.lang.c.]

Fixed...

You're assuming that the code might be C++ because you think it's
doing "overloading". You've misunderstood the meaning of the term.

Overloading means using the same name for two different functions or
operators. In C++, you can have two different functions both named
"foo", as long as they take different types of arguments. When the
compiler sees a call to foo(), it determines which function to call by
looking at the arguments. Likewise, you can overload operators as
functions; the C++ standard library does this to use the shift
operators "<<" and ">>" for I/O. When the compiler sees
(cout << "hello"), it decides what to do based on the types of the
operands.

None of this is supported in C, and none of this has anything to do
with the code you posted.

The code you posted is simply converting an expression of one type
(dummy, a pointer type) to another type (long). This is perfectly
legal in C (and in C++), but the result won't necessarily be
meaningful.

You can overload a "cast" operator in C++. This must be what he thinks
is happening. However, it won't work with the construct the OP is
saying is being worked with. You can't overload the casting of a
primitive type - any kind of real pointer qualifies as such (you can
have things that /look/ like pointers that could be overridden but not
a pointer like in the OP's post).

Example:

struct X
{
public:

operator long () { return 5; }
};

X x;

long y = x;

But this still won't work:

X *x = new X;

long y = (long)x;

Still have ptr value of x.

What the OP thinks is happening is not possible in either language.