Re: [Boost-users] [fusion] transform a heterogeneous vector

Subject: Re: [Boost-users] [fusion] transform a heterogeneous vector
From: Christopher Schmidt (mr.chr.schmidt_at_[hidden])
Date: 2009-09-20 17:20:24

Comments inline ...

Jean-Louis Leroy schrieb:
> 
> First there's the warning :
> 
>    1>c:\program
>    
> files\boost\boost_1_39\boost\fusion\view\transform_view\detail\deref_impl.hpp(42) 
> 
>    : warning C4172: returning address of local variable or temporary
>    1>        c:\program
>    
>    etc etc etc
I assume you return the plain argument type (that is a reference) in 
your triple2::result meta-function. If that's the case, simply wrap the 
argument in a remove_reference.
E.g.

struct triple2
{
        //...

        template<typename Self, typename Arg>
        struct result<Self(Arg)>:
                boost::remove_reference<Arg>
        {};
};

> Also, I see that transform returns a view. I need a proper vector that 
> is distinct from the input vector. Here's a bit of context. I am 
> building a recipient for one row in a database query.  I have a vector 
> of Exprs, one for each column in the select. Expr has a type member. I 
> want to make a vector with an element for each Expr, initialized to the 
> type's default value (as per type()).
> 
> I.e. something like this :
> 
>    struct f { /* ... */ template<typename Expr> operator ()(Expr) { 
> return typename Expr::type(); }
>    typedef result_of::transform<SelectList, f>::type row_type;
>    row_type row;
>    transform(select.exprs, row); // type of select.exprs is SelectList
>    std::deque<row_type> results;
>    while (data) {
>       for_each(row, read_fun(...));
>       results.push_back(row);
>    }
> 
> Alas row_type is going to be a view :-/
> 
simply convert the view (type!) back to a vector using 
fusion::result_of::as_vector.

using namespace boost::fusion;
result_of::as_vector<result_of::transform<SelectList,f>::type>::type vec;

-Christopher