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From: Derek.Ledbetter@giganews.com
Newsgroups: comp.std.c++
Subject: Re: Proposal: Allow member typedefs to be used in return types without qualification
Date: Mon, 25 Apr 2005 15:19:19 GMT
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On 2005-04-19 19:58:06 -0700, jfa1@cec.wustl.edu ("James Aguilar") said:

> Currently, when writing a member function in a class definition, 
> typedefs that the programmer makes in the class allow member typedefs 
> to be used as a return type, thus:
> 
> class Example
> {
> public:
>   typedef bool result_type;
>   typedef int input_type;
> 
>   result_type operator ()(input_type in1, input_type in2)
>   {
>     return in1 < in2;
>   }
> };
> 
> However, when defining a member function outside of the class in which 
> it is declared, it looks more like this:
> 
> Example::result_type Example::operator ()(input_type in1, input_type in2)
> {
>   return in1 < in2;
> }
> 
> and
> 
> result_type Example::operator ()(input_type in1, input_type in2)
> {
>   return in1 < in2;
> }
> 
> is illegal.  This should not be.  If the parameter types can be assumed 
> to be members of the containing class, so should return types.

The "class namespaces" proposal would solve this problem and more, in a 
simpler way.  Here's how it would work:

namespace class Example
{
  result_type operator ()(input_type in1, input_type in2)
  {
    return in1 < in2;
  }
}

See http://www.open-std.org/jtc1/sc22/WG21/docs/papers/2003/n1420.pdf

--
Derek Ledbetter
derekl@serve.com

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