Question

我有一个使用boost :: get of boost：variant模块的模板方法：

typedef boost::variant<int, std::string, bool, uint8_t> Variant;

template <class T>
void write(const Variant& t) {
    size_t sizeT = boost::apply_visitor(SizeOfVisitor(), t);
    memcpy(&v[offset], &boost::get<T>(t), sizeT);
}

问题是我只在运行时知道Variant的基础类型。和AFAIK我只能用which()方法查询它。

有什么方法可以传递类型T，这是此方法的Variant的基础类型？例如，使用which()我知道，有什么类型，但如何传递它？

switch (m_var.which()) { // Returns an int index of types in the order of passed template classes
    case 0: // This is int
    case 1: // This is std::string
    ...
}
...
Writer.write<???>(m_var); // How to pass the type here?

编辑：如果您知道任何其他方法来实现所需的结果 - 实际获取boost :: variant内部变量的地址，从那里复制，请与我分享您的想法

谢谢

Answer 1

我碰巧在这里写了一个非常相似的答案：

的 C++: Extract size of boost::variant element

同样，最重要的是， 完全虚假 将memcpy与非POD数据类型一起使用（所以不能与std::string一起使用。永远。

使用仅在运行时知道的类型对variant进行操作的方法是使用boost::static_visitor<>。

这里的main()示例适合接近您想要实现的目标，显然，

<强> Live On Coliru

#include <boost/variant.hpp> #include <boost/bind.hpp> #include <boost/array.hpp> // just as a sample #include <iostream> namespace serialization { namespace customization { template<typename T, typename Out, typename R = typename boost::enable_if<boost::is_pod<T>, void>::type> void do_serialize(T const& x, Out& out) { static_assert(boost::is_pod<T>(), ""); char const* rawp = reinterpret_cast<char const*>(&x); std::copy(rawp, rawp+sizeof(T), out); } template<typename Out> void do_serialize(std::string const& x, Out& out) { do_serialize(x.size(), out); for(auto ch : x) do_serialize(ch, out); } } struct serialize_f : public boost::static_visitor<> { template<typename Out, typename... T> void operator()(boost::variant<T...> const& v, Out& out) const { boost::apply_visitor(boost::bind(*this, _1, boost::ref(out)), v); } template<typename T, typename Out> void operator()(T const& x, Out& out) const { using customization::do_serialize; // ADL dispatch do_serialize(x, out); } }; template <typename T, typename Out> Out serialize(T const& v, Out out) { const static serialize_f _vis {}; _vis(v, out); return out; } } namespace MyUserTypes { struct A { std::string name; int i; }; template<typename Out> void do_serialize(A const& v, Out& out) { // ADL will find this serialization::serialize(v.name, out); serialization::serialize(v.i, out); } } int main() { using namespace serialization; std::vector<uint8_t> binary_data; auto out_inserter = back_inserter(binary_data); // variants and custom types typedef boost::variant<MyUserTypes::A, boost::array<char, 42> > V; MyUserTypes::A myA { "0123456789", 99 }; V v = boost::array<char,42>(); serialize(myA, out_inserter); serialize(v, out_inserter); v = myA; serialize(v, out_inserter); std::cout << "Bytes in binary_data vector: " << binary_data.size() << "\n"; }

Answer 2

switch (m_var.which()) { // Returns an int index of types in the order of passed template classes
    case 0: // This is int
      Writer.write<int>(m_var);
    break;
    case 1: // This is std::string
      Writer.write<std::string>(m_var);
    break;
    ...
}

：）

如何将类模板参数传递给boost :: variant？

2 个答案: