Re: [boost] [parameter] Preventing instantiation of lazy binding operator() until default is needed

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