Re: [boost] [parameter] Binding type.

From: David Abrahams (dave_at_[hidden])
Date: 2006-03-06 20:39:43

Rene Rivera <grafik.list_at_[hidden]> writes:

> Daniel Wallin wrote:
>> 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.
>
> OK I understand what you are saying, but I don't see it :-) As far as I 
> understand from the docs the binding<>::type has nothing to do with the 
> default value as above as it happens before one gets to extracting the 
> value. I wasn't talking about changing the types that the operator[] 
> return. 

Consider the usage you're proposing when you /do/ want a reference to
bind to the actual argument:

  typename binding<Args,tag::k,int>::type const& x = args[k | 0]

If the "k" argument wasn't supplied, your int const& will be a
dangling reference.

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com