On Sat, 2004-11-27 at 09:24, Hendrik Tews wrote:
> Jacques Garrigue <garrigue@math.nagoya-u.ac.jp> writes:
> 
>    Note that void in C is definitely not zero. You cannot have variables
> 
> C++ standard, 3.9.1.9: The void type has an empty set of values. ...
> 
> So I would say void is zero. On the other side you have functions
> returning void. Therefore I would conclude that the type theory
> of C++ is unsound.

I'm not sure I quite understand (not that I disagree with 
the conclusion..)

void in C++ is initial. However, this function:

	int f();

is NOT a function

	void -> int

because in C/C++ you have a list of arguments:

	g(a,b,c)

If you construe the list as a product, so that the function

	int g(int,float,long)

has type

	int * float * long -> int

then you must construe the type of f as

	unit -> int

since the product of an empty list is unit, and NOT void.
The notation for application:

	f ()

even looks like f is being applied to an Ocaml empty tuple :)

In particular in C++ you will note that this does NOT
type correctly:

	void f();
	f( f() );

because the type of f() is void, and the type of the argument
is actually unit.

This means C/C++ has stronger typing in this respect
than Ocaml :)

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