🚀 go-pugleaf

In article <1992Nov11.092316.306@aut.abb.se>, dlarsson@aut.abb.se (Daniel Larsson) writes:
> To my surprise, I noticed that operator -> can only return the following
> three types:
>    - a pointer to a class,
>    - a class, or
>    - a reference to a class.
> 
> This means I cannot do simple things like:
> 
>   template< class T >
>   class OffsetPtr 
>     {
>     public:
>       OffsetPtr();
>       OffsetPtr( T* );
>       ...
>       T* operator ->();
>       ...
>     };
> 
>   OffsetPtr<char> ptrToString;
> 
> What is the rationale behind this restriction? I couldn't find anything in
> C++PL 2nd ed. that mentions this restriction, although I didn't look too hard.

a->b
  ...where a is of a class type is interpreted as...
(a.operator->())->b
  ...which in turn may expand to...
((a.operator->()).operator->())->b
  ...and so on.

but in all cases the thing on the rhs of the -> is a class member. Since
types other that structs and classes do not have members it would be
useless for operator->() to return anything other than a pointer to a class or
a class for which operator->() is also defined.

Now what you could do is define operator T* (). Unfortuneately if you write
`a->b' where a is of a class without an operator->() the compiler does not
try compiling it as `(*a).b' so you'll have to write this out in full each
time. Likewise you can't write `a[5]' but you can write `*(a+5)'.

This has annoyed me in the past and I feel that the language definition
should employ user defined type conversions to resolve the lh operand of
the -> ->* and [] operators.
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