David Abrahams wrote:
Now I just ran into another issue, from the other side of the result_of interface. If I have a function template taking a function object argument by reference, and I want to use result_of with it, I need to do something special to account for the case where I've been passed a function reference. To wit:
template <class F> typename result_of<F()>::type call(F const& f) { return f(); }
int f() { return 0; }
int x = call(f);
The problem is that in the above case, F is deduced as int(), and we form an illegal function type (one that itself returns a function type) in result_of<F()>::type.
AFAIK, there is another defect. This doesn't compile: template<class F> typename result_of<typename result_of<F()>::type()>::type call_call(F f) { return f()(); } if `f()` returns a lvalue FunctionObject.
To deal with this I need something more like
template <class F> typename result_of< typename mpl::if_< is_class<F> , F , typename add_pointer<F>::type >::type() >::type call(F const& f) { return f(); }
Is there an easier way?
How about `boost::decay`? Or an "imaginary" function `apply` might be better. template<class F> typename result_of<typeof_apply(F const &)>::type call(F const &f) { return apply(f); } Regards, -- Shunsuke Sogame