> On most C syntax I do just fine, but function pointers are always a
> head-scratcher for me.

Yes, it can be a real pain...

>=20
> If I wanted to declare a pointer to this function:
>=20
> int foo(int bar);
>=20
> What would be the syntax for the declaration?
>=20
> The best I came up with was this:
>=20
> (int)(*)(int) pfoo =3D &foo;
>=20

Try:

int (*pfoo)(int bar)=3Dfoo;=09

You don't need to do &foo.

If you want to declare an array of function pointers you have=20
to do it like this:

int (*pfunarray[])(int bar)=3D{
        one_fun,
        another_fun,
        andsoon_fun,
        NULL
};

The NULL is only needed if you dynamically want to scan the=20
array and like to know where it ends.

Note that int C++ you can only get a pointer to a class member=20
function if it is declared as static (because of the this=20
pointer that otherwise also would be needed).

Regards / Ruben

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D
Ruben J=F6nsson
AB Liros Electronic
Box 9124, 200 39 Malm=F6, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
ruben@pp.sbbs.se
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D

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