Re: How to disable incorrect conversion ?

[Ruslan Shevchenko <Ruslan at Shevchenko dot Kiev dot UA>]

Martin von Loewis wrote:

> > If you would  look on this problem, please, contact with me.
>
> Ruslan,
>
> Here is a stripped-down version of the code you sent me which still
> produces a warning:
>
> class CORBA_String_var
> {
>     char* ptr_;
>
> public:
>     operator char*&() { return ptr_; }
>     operator const char*() const { return ptr_; }
> };
>
>
> void message(const char*);
>
> void f()
> {
>     CORBA_String_var s;
>     message(s);
> }
>
> v.cc: In function `void f()':
> v.cc:17: warning: choosing `CORBA_String_var::operator char *&()' over `CORBA_String_var::operator const char *() const'
> v.cc:17: warning:   for conversion from `CORBA_String_var' to `const char *'
> v.cc:17: warning:   because conversion sequence for `this' argument is better
>
> Is this the problem you are talking about? If so, the compiler is right:
> This call expands to
>
>    char*& tmp = s.operator char*& ();
>    message ((char*)tmp);
>
> This is a user-defined conversion sequence. Therefore, the compiler
> must do overload resolution for the conversion candidates. There are
> two candidates, one expecting a "CORBA_String_var&", the other one
> "const CORBA_String_var&", and the actual argument (i.e. *this) is
> of lvalue type CORBA_String_var.
>
> The identity conversion of this is better than the qualification
> conversion in this case, so the compiler choses (rightly) the
> non-const converter. It then ends up with a char*&, which gets
> standard-converted to const char* as an argument to message.

   Thanks. I understood.   Yet another  questions:

       If I have hypotetic situation with 2 operators:

       operator char*  () const {  return new_sz(_ptr); }
       operator const char* ()  const { return   _ptr;  }

       And I cand belive, that  the first operator will not be called
        (during passing to  message(const char*) )

       why to produce warning ?

         It means, that I can't  pass  parameter without warning, when I
         have few conversion operators  without casting ?


> If you want the const version to be called, you'd have to make s a
> const object. I cannot see why you would want to do this, though,
> since both conversion operators have identical implementations.
>

  It is not my program ;)  It is a big package, which was developed during few years.

> Hope this helps,
> Martin

   Can you full compile the code, which I  was send you ?

  My egcs (2.91.29)  is hangs
                    (in Utils.cpp, line 85).  (I test this before sending you .  )

WBR.

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