Re: [Boost-users] [xpressive] Using placeholder in semantic actions

Subject: Re: [Boost-users] [xpressive] Using placeholder in semantic actions
From: Eric Niebler (eric_at_[hidden])
Date: 2008-09-04 12:58:08

Symons, Greg - SSD wrote:
> Hi!
> 
> I'm trying to use xpressive with semantic actions that use a placeholder, but I
> can't quite see the right way to do it. Here's my code:
> 
> namespace grammar
> {
>    using namespace boost::xpressive;
> 
>    struct ParsedUri
>    {
>       std::string hostname;
>       std::string modelPath;
>       std::string diagramPath;
>    };
> 
>    static placeholder<ParsedUri> result;
<snip>

>    static sregex              pathHost       =
>                                  (host= hostname) >> pathSep >> path
>                                  [ result.hostname = host ];
>    static sregex              modelPath      = pathNoHost | pathHost
>    static sregex              uri            =
>                                  "rosert://" >> (model= modelPath) >> '?' >> (diagram= path)
>                                  [ result.modelPath = model, result.diagramPath = diagram ];
> }
> 
> What happens is that when I compile the code, I get a compile error for the
> lines accessing the placeholder saying that the various members of the structure
> are not members of placeholder<T>, which looking at regex_actions.hpp seems to
> make sense to me. 

Right.

> But in the example using placeholder<T> in the docs, it looks
> like you can use it in the semantic action as if it were a T. Somewhere I'm
> misunderstanding something, and I don't know what. Can anyone help me here?

Xpressive overloads operators to give the impression that your can use a 
placeholder<T> as a T. But xpressive cannot overload operator. to give 
the impression of data member access because in C++ you can't overload 
operator..

Try this instead...

   xpressive::function<std::string ParsedUri::*>::type getHostname =
       {&ParsedUri::hostname};

Now in your semantic actions, instead of doing this:

   [ result.hostname = host ]

You can do this:

   [ result->*getHostname() = host ]

Xpressive overloads operator->*, which is the closest thing to operator..

Note to self: it would be better if we could leave off the parens in the 
above, like this: [ result->*getHostname = host ]. There's always more 
to do.

HTH,

-- 
Eric Niebler
BoostPro Computing
http://www.boostpro.com