Re: comparing function pointer with int does not produce error

["Martin v. Loewis" <martin@v.loewis.de>]

> On Tue, Oct 23, 2001 at 08:21:26PM +0200, Martin v. Loewis wrote:
> > > 	The following code does not produce an error when one is expected.
> > > void stat ();
> > > 
> > > static const int success = 0;
> > > 
> > > void monk ()
> > > {
> > >   if (stat == success);
> > > };
> > 
> > Thanks for your report. This is not a bug in the compiler: both the
> > function pointer and the integer convert to bool (4.12). So this is a
> > boolean comparison, and thus well-formed C++.
> 
> Really?

You are right, I was confused (I thought of the operator signatures in
13.6/12, ignoring that those aren't used if none of the arguments was
of class type).

I'll take another attempt at declaring gcc behaviour to be correct.
According to 4.10/1, "success" is a null pointer constant:

# A null pointer constant is an integral constant expression (5.19)
# rvalue of integer type that evaluates to zero. A null pointer
# constant can be converted to a pointer type; the result is the null
# pointer value of that type and is distinguishable from every other
# value of pointer to object or pointer to function type.

5.19/1 defines "integral constant expressions":

# An integral constant-expression can involve only literals (2.13),
# enumerators, const variables or static data members of integral or
# enumeration types initialized with constant expressions (8.5),
# non-type template parameters of integral or enumeration types, and
# sizeof expressions.

Since "success" is an integral constant expression of integer type, it
is a null pointer constant, and can be converted to void(*)().  So
this is a function pointer comparison, and thus well-formed C++ :-)

Regards,
Martin