Re: [Boost-users] [fusion] remove_if and as_vector with Placeholder Expression as predicate

Subject: Re: [Boost-users] [fusion] remove_if and as_vector with Placeholder Expression as predicate
From: Philipp Schwaha (philipp_at_[hidden])
Date: 2012-11-12 16:53:52

hi Felipe!

On 11/12/2012 06:11 PM, Felipe Magno de Almeida wrote:
> Hello,
> 
> I'm trying to remove_if a few elements from a fusion sequence. But it
> is instantiating
> my predicate with the placeholder from MPL later on. Including with as_vector.
> 
> The following code generates the error code I'm pasting by the end of
> the message.
> I'm using GCC 4.7.2 and Linux x86_64. If I wrap meta_func<mpl::_1> in
> a mpl::lambda<>::type
> then the errors following the first error messages is what I get. I'm
> not sure if I'm doing
> something wrong here. If anyone could please shed some light, I would
> really appreciate it.
> 
> #include <boost/fusion/include/as_vector.hpp>
> #include <boost/fusion/include/remove_if.hpp>
> 
> namespace mpl = boost::mpl;
> namespace fusion = boost::fusion;
> 
> struct element
> {
>   typedef mpl::false_ type;
> };
> 
> template <typename T>
> struct meta_func
> {
>   typedef typename T::type type;
> };
> 
> int main()
> {
>   fusion::vector<element, element> e;
>   fusion::as_vector(fusion::remove_if<meta_func<mpl::_1> >(e));
> }
> 
a simple solution would be to make meta_func of the form:

template<typename T>
struct meta_func
{
   typedef mpl::true_ type ;
};

template<>
struct meta_func<element>
{
   typedef mpl::false_ type ;
};

but I'm not sure if it's the way you intend it, since remove_if will
remove items when the predicate evaluates to true, so maybe swap the logic?

hope it helps ;)

cheers
        Philipp