[boost] Re: [typeof][result_of] Separated at Birth?

From: David Abrahams (dave_at_[hidden])
Date: 2005-06-09 14:00:00

jarvi <jarvi_at_[hidden]> writes:

> On Jun 9, 2005, at 12:18 PM, Douglas Gregor wrote:
>
>>
>> On Jun 9, 2005, at 11:49 AM, David Abrahams wrote:
>>> It seems to me that when BOOST_TYPEOF becomes available,
>>> boost::result_of should use it in its default implementation,
>>> something like:
>>>
>>>   template <class F, class A>
>>>   struct result_of<F(A)>
>>>   {
>>>       typedef typename BOOST_TYPEOF(make<F>() ( make<A>() ) ) type;
>>>   };
>>>
>>> Is that in the plan?
>>
>> Absolutely.
>>
> The TR spec says that the implementation can use whatever means to 
> determine the member type type that produces
> the exact type, and if it cannot determine the exact type, then the 
> specs
> details out its behavior:
>
>
>   1   If F is a function type, type is the return type of the function 
> type F.
>   2   If F is a member function type, type is the return type of the 
> member function type F.

These two are redundant.  That *is* the exact type, so the
implementation can determine it.

>   3   If F is a function object defined by the standard library, the 
> method of determining type is unspecified.
>   4   If F is a class type with a member type result_type, type is 
> F::result_type.
>   5   If F is a class type with no member named result_type or if 
> F::result_type is not a type:
>               a   If N=0 (no arguments), type is void.
>            b  If N>0, type is F::result<F(T1, T2, ..., TN)>::type.
>
>   6   Otherwise, the program is ill-formed.

Seems to me that the whole behavior we need for the primary
boost::result_of template is:

1   If F is a function type, type is the return type of the function
    type F.

2   If F is a member function type, type is the return type of the
    member function type F. 

3   If BOOST_TYPEOF knows the result type of the expression, that is
    type.  Note that we're on slightly shaky ground here because we'll
    mis-report const lvalues as rvalues... unless someone can figure
    out how to apply Eric N.'s ?: discoveries to know for sure.

The above steps constitute "using whatever means to determine the
exact type."

4   If F is a class type with a member type result_type, type is
    F::result_type. 

5 	If F is a class type with no member named result_type or if
    F::result_type is not a type:

    a  If N=0 (no arguments), type is void.
    b  If N>0, type is F::result<F(T1, T2, ..., TN)>::type.

6   Otherwise, the program is ill-formed.

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