Semantics operator->()

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2 total messages Started by Douglas C. Schmidt Thu, 24 Aug 1995 03:56

#1833

Author: Douglas C. Schmidt
Date: Thu, 24 Aug 1995 03:56

59 lines
1984 bytes

Hi,

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

	For example, consider the following:
	
----------------------------------------
#include <stdio.h>

class Foo
{
public:
  Foo (int i): i_ (i) {}

  int *operator->()
  {
    return &this->i_;
  }

private:
  int i_;
};

int main ()
{
  Foo foo (10);
  printf ("%d, ", ++*(int *) foo.operator->());
  printf ("%d\n", ++*(int *) foo.operator->());
  // prints "11, 12"
  return 0;
}
----------------------------------------

	When I compile this with gcc it compiles and runs fine.
However, when I use SunC++ 4.0.1 I get the following error:

----------------------------------------
% CC foo.C
"foo.C", line 9: Error: Cannot have a return type of int* for Foo::operator->().
1 Error(s) detected.
----------------------------------------

	I tried to find a discussion of this in the ARM, but couldn't
find it discussed there.  Can someone please tell me if this is legal
in the draft ANSI/ISO C++ standard?

	Thanks,
	
		Doug
-- 
Dr. Douglas C. Schmidt 			(schmidt@cs.wustl.edu)
Department of Computer Science, Washington University
St. Louis, MO 63130. Work #: (314) 935-7538; FAX #: (314) 935-7302
http://www.cs.wustl.edu/~schmidt/

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Re: Semantics operator->()

#1847

Author: Joel Bender
Date: Fri, 25 Aug 1995 02:19

43 lines
1715 bytes

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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