jarvi <jarvi@cs.tamu.edu> 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