Re: [boost] [parameter] Binding type.

From: Daniel Wallin (dalwan01_at_[hidden])
Date: 2006-03-06 17:02:26

Rene Rivera wrote:
> I've been Boost.Parameter a bit by now and I have some usability 
> feedback. There is one thing that currently bugs be a bit, the type 
> defined by parameter::binding. All my uses so far have been of the form:
> 
> typedef typename
>    boost::remove_reference<
>      boost::remove_const<
>        boost::parameter::binding<ArgPack,tag::something>::type
>        >::type
>      >::type
>    something_t;
> 
> something_t s = args[something];
> 
> Or equivalent thereof. So every use I have I end up removing the ref and 
> const because I'm really interested in the original type, and hence want 
> to make a copy. Now I understand the rational outlined in "Eliminating 
> Copies" 
> <http://www.boost.org/libs/parameter/doc/html/index.html#eliminating-copies>. 
> But I think that isn't enough of a justification for dirtying up the 
> user interface, as above. I'd rather that the binding type where the 
> value type so that one would have:
> 
> typedef typename
>    boost::parameter::binding<ArgPack,tag::something>::type
>    something_t;
> 
> something_t s = args[something];
> 
> Having it that way one still has the option of more naturally using the 
> reference type as;
> 
> something_t const & q = args[something];

I don't think the rationale for this is completely expressed in the
docs. The reason binding<> works like this is because of dangling
references to defaults. Consider from your example above:

  something_t const & q = args[something | 0];

When binding<> returns the default type here you will get a dangling
reference here:

  int const& = args[something | 0]

Our design handles this by letting binding<> return a reference type
when an argument is bound to the keyword, and otherwise return the
default type unchanged.

I would consider adding a binding_value<> metafunction though.

-- 
Daniel Wallin