问题
Assume that you have a boost::any
object and a boost::variant
object.
I'm looking for a generic function convert
, that takes a template parameter T being a specialized boost::variant
e.g. boost::variant<int, std::string>
and magically converts the boost::any
to one of the available types of the given boost::variant
.
template<T>
T convert(const boost::any& any) {
// Some generic conversion code here or throw exception if conversion is not possible!
}
int main(int argc, char** args) {
typedef boost::variant<int, std::string> TVar;
boost::any any="Hello World";
TVar variant=convert<TVar>(any);
// variant contains "Hello World"
return 0;
}
I'm wondering if it is possible to write such a function or if it might be impossible for some reason?
回答1:
Let's enclose all code in struct templated by variant type
template<class VAR>
struct ExtractorGenerator
{
using Extractor = std::function<boost::optional<VAR>(boost::any const &)>;
std::vector<Extractor> extractors;
template<class T>
static boost::optional<VAR> tryCast(boost::any const & arg);
template<class T>
void operator()(T);
};
You can easily write a function that for given type tries to convert boost::any to variant of this type
template<class VAR>
template<class T>
boost::optional<VAR> ExtractorGenerator<VAR>::tryCast(boost::any const & arg)
{
T const * val = boost::any_cast<T>(&arg);
return val == nullptr ? boost::none : boost::make_optional(VAR{*val});
}
Now using boost::mpl you can iterate through all variant types to generate function for each variant's type
template<class VAR>
template<class T> void ExtractorGenerator<VAR>::operator()(T)
{
extractors.push_back(Extractor::tryCast<T>);
}
typedef boost::variant<int, std::string, char> MyVariant;
ExtractorGenerator<MyVariant> generator;
boost::mpl::for_each<MyVariant::types>(boost::ref(generator));
And now you just apply all created functions:
std::vector<MyVariant> extractedVals;
for (auto fun : extractor.extractors)
{
boost::optional<MyVariant> extracted = fun(val);
if (extracted)
extractedVals.push_back(extracted.get());
}
回答2:
You can call boost::any_cast
for each of the types within boost::variant
and stop when the first cast succeeded:
#include <iostream>
#include <utility>
#include <stdexcept>
#include <sstream>
#include <boost/any.hpp>
#include <boost/variant.hpp>
#include <boost/type_index.hpp>
#include <boost/mpl/size.hpp>
#include <boost/mpl/at.hpp>
template <typename Sequence>
struct mpl_sequence_to_std_tuple
{
template <std::size_t... Is>
static auto x(std::index_sequence<Is...>) -> std::tuple<typename boost::mpl::at_c<Sequence, Is>::type...>;
using type = decltype(x(std::make_index_sequence<boost::mpl::size<Sequence>::type::value>{}));
};
struct signal_conversion_success{};
template <typename T, typename Variant>
void try_convert(const boost::any& any, Variant& var)
{
try
{
var = boost::any_cast<T>(any);
throw signal_conversion_success{};
}
catch(const boost::bad_any_cast &)
{
}
}
template <typename T, typename... Ts>
std::string parameter_pack_to_string(const std::string& separator = ", ")
{
std::stringstream ss;
ss << boost::typeindex::type_id<T>().pretty_name();
auto l = {0, (void(ss << separator << boost::typeindex::type_id<Ts>().pretty_name()),0)...};
std::ignore = l;
return ss.str();
}
template <typename Variant, typename...Ts>
void do_convert(const boost::any& any, Variant& var, std::tuple<Ts...>)
{
bool success = false;
try {
auto l = {0, (void(try_convert<Ts>(any, var)), 0)... };
std::ignore = l;
}
catch(const signal_conversion_success&)
{
success = true;
}
if (!success)
{
std::stringstream ss;
ss << "cannot convert this boost::any instance to any of the following types: ";
ss << parameter_pack_to_string<Ts...>();
throw std::invalid_argument(ss.str());
}
}
template<typename Variant>
void convert(const boost::any& any, Variant& var)
{
using Tuple = typename mpl_sequence_to_std_tuple<typename Variant::types>::type;
do_convert(any, var, Tuple{});
}
struct print_visitor : public boost::static_visitor<void>
{
template <typename T>
void operator()(T&& t) const
{
std::cout << boost::typeindex::type_id<T>().pretty_name() << ": " << std::forward<T>(t) << std::endl;
}
};
int main()
{
using Variant = boost::variant<int, std::string>;
boost::any any = std::string("Hello World");
Variant var;
convert(any, var);
boost::apply_visitor(print_visitor(), var);
}
live example
In case none of the casts succeeded, an exception is thrown, see the following live example.
来源:https://stackoverflow.com/questions/35357614/generic-function-to-convert-boostany-to-boostvariant