Extend boost property tree get for own types

Question

How do I extend boost property tree to support my own types?

e.g. I want to write

boost::property_tree::basic_ptree<std::string,std::string> xml;
...
xml.get<int>("level"),
xml.get<double>("accuracy"),
xml.get<Ship>("sauce_pan"),
...

And then I guess somewhere I'd like to describe how to get a Ship from a boost::property_tree::basic_ptree<std::string,std::string> or something similar.

pmr · Accepted Answer · 2014-06-15T08:42:16.040

You need to provide a default constructor for Ship and a stream input operator:

struct Ship {
  friend std::istream& operator>>(std::istream& s, Ship& e) {
    /* read ship data from s */
    return s;
  }
};

Unfortunately, I'm not sure if this is an official feature of property_tree as I can not find it in the documentation.

For more fine-grained access get also takes a Translator as it's second argument.

struct ShipTranslatorJSON {
  // the type we return
  typedef Ship external_type;
  // the type expect as an argument
  typedef std::string internal_type;

  boost::optional< external_type > get_value(const internal_type&) {
    return external_type();
  }
  // only required when you actually use put
  // boost::optional< T > put_value(const T &); 
};


// use as
xml.get<Ship>("sauce_pan", ShipTranslatorJSON());

This method also seems more "official", although nothing in the documentation actually explains what a Translator really is.

And how do I specialize that according to `JSON` or `xml`? – Cookie Jun 15 '14 at 08:29 — Cookie, Jun 15 '14 at 08:29
@Cookie See edit using `Translator` argument for `get`. – pmr Jun 15 '14 at 08:42 — pmr, Jun 15 '14 at 08:42

Extend boost property tree get for own types

1 Answers1