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