🚀 go-pugleaf

kristofer stenberg <kristofer.stenberg@chello.se> wrote in message
news:daLb5.765$u21.189364@nntp1.chello.se...
> Say I have an object Dog and a method Dog::run().
> ptr is a pointer to Dog and funcPtr is pointer to run(). First I try to
call
> run() with (ptr ->*   funcPtr) () and it works fine. Then I try (ptr ->
> *funcPtr) ()
> and its an error. A space between -> and * shouldnt make a difference,
> should it?

Yes it does.  ->* is a *single* operator (pointer-to-member dereference,
via a pointer to the type having the member).  Putting a space between
-> and * means they're separate operators, in which case, their being
adjacent is syntactically wrong.

Also, .* is a *single* operator (pointer-to-member dereference,
via an object of the type having the member)

Dog fido;
Dog *pf = &fido;
fido.*funcPtr();
pf->*funcPtr();

With the statement:
pf-> *funcPtr();

pf-> means I want to select a particular member (by name),
and '*funcPtr' certainly isn't one, so you'll probably get a
syntax error.

-Mike