這是絕對有可能的(見下文),但我不確定它是否值得麻煩...難道你不能使用更簡單的方法,也許是繼承和多態嗎?
儘管如此,這裏是用Boost.MPL一個有效的解決方案:
// MapperRow holds the necessary parsing informations: type of the member,
// type of the object, pointer to the appropriate member, parsing function
template<typename Type, typename Clazz, Type Clazz::*Member,
void (*Parser)(const std::string &, Type &)>
struct MapperRow
{
typedef Type type;
typedef Clazz clazz;
typedef Type Clazz::*memberType;
static const memberType member;
typedef void (*parserType)(const std::string &, Type &);
static const parserType parser;
};
template <typename Type, typename Clazz, Type Clazz::*Member,
void (*Parser)(const std::string &, Type &)>
const typename MapperRow<Type, Clazz, Member, Parser>::memberType
MapperRow<Type, Clazz, Member, Parser>::member = Member;
template <typename Type, typename Clazz, Type Clazz::*Member,
void (*Parser)(const std::string &, Type &)>
const typename MapperRow<Type, Clazz, Member, Parser>::parserType
MapperRow<Type, Clazz, Member, Parser>::parser = Parser;
// fill iterates over a map key->MapperRow, trying to find the given key.
// if found, it calls the parsing function, otherwise it asserts false (should
// probably throw an exception instead)
template <typename Clazz, typename First, typename Last>
struct fill_impl
{
static void apply(Clazz &obj, const std::string &key,
const std::string &value)
{
typedef typename mpl::deref<First>::type entry;
static const char *curKey =
mpl::c_str< typename
mpl::first<entry>::type
>::value;
if (key == curKey)
{
typedef typename mpl::second<entry>::type Row;
Row::parser(value, obj.*Row::member);
}
else
{
fill_impl<
Clazz, typename
mpl::next<First>::type,
Last
>::apply(obj, key, value);
}
}
};
template <typename Clazz, typename Last>
struct fill_impl<Clazz, Last, Last>
{
static void apply(Clazz &obj, const std::string &key,
const std::string &value)
{
assert(false && "key not found");
}
};
template <typename Map, typename Clazz>
void fill(Clazz &obj, const std::string &key, const std::string &value)
{
fill_impl<
Clazz, typename
mpl::begin<Map>::type, typename
mpl::end<Map>::type
>::apply(obj, key, value);
}
使用示例:
template <typename T>
void get_value_units(const std::string &str, T &value)
{
value = T::from_value(boost::lexical_cast<typename T::value_type>(str));
}
template <typename T>
void get_value_simple(const std::string &str, T &value)
{
value = boost::lexical_cast<T>(str);
}
typedef boost::units::quantity<boost::units::si::energy> Energy;
typedef boost::units::quantity<boost::units::si::length> Length;
struct DeathStar
{
Length radius;
Energy energy;
bool operational;
int tie_fighters;
};
// Could be clearer with MPLLIBS_STRING*
typedef mpl::map<
mpl::pair<
mpl::string<'deat','h_st','ar_r','adiu','s'>,
MapperRow< Length , DeathStar, &DeathStar::radius,
&get_value_units<Length> > >,
mpl::pair<
mpl::string<'deat','h_st','ar_e','nerg','y'>,
MapperRow< Energy, DeathStar, &DeathStar::energy,
&get_value_units<Energy> > >,
mpl::pair<
mpl::string<'oper','atio','nal'>,
MapperRow< bool, DeathStar, &DeathStar::operational,
&get_value_simple<bool> > >,
mpl::pair<
mpl::string<'num_','tie_','figh','ters'>,
MapperRow< int, DeathStar, &DeathStar::tie_fighters,
&get_value_simple<int> > >
> death_star_map;
int main()
{
DeathStar ds;
fill<death_star_map>(ds, "death_star_radius", "12");
fill<death_star_map>(ds, "death_star_energy", "34");
fill<death_star_map>(ds, "operational", "1");
fill<death_star_map>(ds, "num_tie_fighters", "56");
std::cout << "Radius: " << ds.radius << '\n';
std::cout << "Energy: " << ds.energy << '\n';
std::cout << "Operational: " << std::boolalpha << ds.operational << '\n';
std::cout << "Tie fighters: " << ds.tie_fighters << '\n';
}
* MPLLIBS_STRING
我不明白的問題,但你可以在模板聲明中不會像使用字符串那樣使用類。這隻適用於像'bool'或'unsigned'這樣的基本類型。 – Peter
可能有mpl包裝類型,它以某種方式嗎?我可能會問不可能的事情,但我已經看到類似於升壓狀態機的東西,所以我希望能夠接近。 – MattSmith
是的,有辦法。不是很難(使用一個有許多字符作爲參數的模板)。爲此,我發現使用代碼生成器可能更簡單,因爲您已經擁有XML,只需要一個xsl或其他類型的轉換器來生成代碼。 – Peter