Subject: Re: [boost] [parameter] Preventing instantiation of lazy binding operator() until default is needed
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2009-05-29 17:17:50
AMDG
Jeremiah Willcock wrote:
> This is a subset of my previous email, with a more specific and
> concrete example. The following code:
>
> #include <boost/parameter/name.hpp>
> #include <boost/parameter/binding.hpp>
>
> BOOST_PARAMETER_NAME(my_param)
>
> template <typename T> struct s {};
>
> template <typename T>
> struct bogus {
> typedef typename s<T>::type result_type;
> template <typename U> struct result: public s<U> {};
> result_type operator()() const {return 0;}
> };
>
> template <typename ArgPack>
> void f(const ArgPack& ap) {
> const char* cs = ap[_my_param || bogus<const char*>()];
> }
>
> int main(int, char**) {
> f(_my_param = "foo");
> return 0;
> }
>
> does not work because result_type is invalid when the struct bogus is
> instantiated when creating the default value used in operator||(). If
> operator()() was a template, I think the compiler would delay
> instantiating it until it was called. Is there a way to force
> operator()() to be a template in Boost.Parameter, allowing its return
> type to only be instantiated when Boost.Parameter knows that the
> default will be used? This is a simplified example from a larger
> piece of code where the lazy default does not have a valid return type
> in some cases, and the parameter must be explicitly specified in those
> cases. Thank you.
In this case, there's nothing Boost.Parameter can do about it, since
result type is instantiated when you
construct bogus<const char*>(). There is no way to make operator() a
template without having
variadic templates or by making Boost.Parameter pass an argument to
operator().
In Christ,
Steven Watanabe