通用功能转换的boost ::任何提振::变种 [英] Generic function to convert boost::any to boost::variant

问题描述

假设你有一个的boost ::任何对象和的boost ::变种对象。

Assume that you have a boost::any object and a boost::variant object.

我在寻找一个泛型函数转换，带有一个模板参数T是一个专门的的boost ::变种例如： 的boost ::变种＆LT; INT，标准::字符串＆GT; 和神奇的的boost ::任何转换为之一可用的类型给定的的boost ::变种。

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;
}

我不知道是否有可能写这样的功能，或者它可能会因为某些原因不可能？

I'm wondering if it is possible to write such a function or if it might be impossible for some reason?

推荐答案

让我们附上所有code的结构由变量类型模板

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);
};

您可以轻松地编写一个函数，给定类型的尝试转换的boost ::任何这种类型的变种

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});
}

现在使用boost :: MPL可以通过所有变量类型迭代生成每个变量的类型函数

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));

和现在你只申请创建的所有功能：

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());
}

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