🚀 go-pugleaf

In article <41gq8b$1jm@tango.cs.wustl.edu> Douglas C. Schmidt  wrote:

>         I've got a question about the behavior of operator->(), in
> particular, can it return a pointer to a builtin type?

I can never seem to find where this is defined in the ARM, but
operator->() can only return something where operator->() is defined. 
Seems kinda strange, but it makes sense.

   struct S { int s; };

   S  *p;

   class C { S *operator->(void); };

   C  c;

In this case "c->s" is OK because "c.operator->()" returns something of type 
pointer to S and you would expect to have "p->s" work.  The way I think
about this now is "p->s" is the same as "p.operator->().s".

So "c->s" is the same as "c.operator->().operator->().s".

   int *px;

So *px is meaningful, but px-> is not defined, nor is "px.operator->()".


Note that "px.operator*()" IS defined, so you could do something like this:

   class Q { int& operator*(void); };

   Q q;

Now "*q" is an integer, the same one as refered to by "q.operator*()".

Does this help?

Joel
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