我正在尝试使用boost :: any来封装sqlite返回值。然后我尝试编写一个循环来打印这些。
我的第一个想法是做类似的事情:
for(boost::any field: row) {
switch(field.type()) {
case typeid(double):
double value = any_cast<double>(field);
...
break;
case typeid(other type):
...
}
}
现在对于有经验的程序员来说,显然这不起作用,因为typeid返回实例而不是数字id。
经过一些研究后,我想我可能会尝试typeid(...).hash_code()
,但这不足以constexpr
限定(除了哈希冲突的危险)。
if ... else ...
迷宫更好的方法来处理基于typeid的对象?hash_code
不是const_expr
的原因是什么?这是单独编译目标文件的结果吗?std::type_index
有什么用?考虑到它只提供了一些额外的运算符(<
,<=
,>
,>=
),为什么无法将其功能与std::type_info
集成?< / LI>
醇>
答案 0 :(得分:7)
我有一种感觉,你正在寻找提升变量和静态访问。
由于没有提到变体,这可能值得作为答案发布。 Demonstruction:
<强> Live On Coliru 强>
#include <sstream>
#include <iostream>
#include <boost/variant.hpp>
using namespace boost;
struct Nil {};
using blob_t = std::vector<uint8_t>;
using field_value_t = boost::variant<Nil, double, char const*, long, blob_t/*, boost::date_time, std::vector<uint8_t>*/>;
struct handler : static_visitor<std::string> {
std::string operator()(double) const { return "double"; }
std::string operator()(char const*) const { return "C string (ew!)"; }
std::string operator()(long) const { return "long"; }
std::string operator()(blob_t) const { return "long"; }
std::string operator()(Nil) const { return "<NIL>"; }
template<typename T>
std::string operator()(T const&) const { throw "Not implemented"; } // TODO proper exception
};
void handle_field(field_value_t const& value) {
std::cout << "It's a " << apply_visitor(handler(), value) << "\n";
}
int main() {
handle_field({});
handle_field(blob_t { 1,2,3 });
handle_field("Hello world");
handle_field(3.14);
}
打印
It's a <NIL>
It's a long
It's a C string (ew!)
It's a double
答案 1 :(得分:5)
这是使用C ++ 11 lambdas在boost::any
上类似于静态访问的实现:
#include <iostream>
#include <type_traits>
#include <boost/any.hpp>
template <size_t, typename...>
struct select_type { };
template <size_t index, typename First, typename... Types>
struct select_type<index, First, Types...> : public select_type<index - 1, Types...> { };
template <typename First, typename... Types>
struct select_type<0, First, Types...>
{
using type = First;
};
template <typename T>
struct function_traits : public function_traits<decltype(&T::operator())> { };
template <typename Return, typename Class, typename... Args>
struct function_traits<Return (Class::*)(Args...) const>
{
using result_type = Return;
template <size_t argN>
using argument_type = select_type<argN, Args...>;
};
template <typename... Functors>
struct any_call_impl
{
static bool call(boost::any &, Functors const & ...)
{
return false;
}
static bool call(boost::any const &, Functors const & ...)
{
return false;
}
};
template <typename FirstFunctor, typename... Functors>
struct any_call_impl<FirstFunctor, Functors...>
{
static bool call(boost::any & v, FirstFunctor const & first, Functors const & ... rest)
{
using arg = typename function_traits<FirstFunctor>::template argument_type<0>::type;
using arg_bare = typename std::remove_cv<typename std::remove_reference<arg>::type>::type;
if (v.type() == typeid(arg_bare)) {
first(*boost::any_cast<arg_bare>(&v));
return true;
}
return any_call_impl<Functors...>::call(v, rest...);
}
static bool call(boost::any const & v, FirstFunctor const & first, Functors const & ... rest)
{
using arg = typename function_traits<FirstFunctor>::template argument_type<0>::type;
using arg_bare = typename std::remove_cv<typename std::remove_reference<arg>::type>::type;
if (v.type() == typeid(arg_bare)) {
first(*boost::any_cast<arg_bare>(&v));
return true;
}
return any_call_impl<Functors...>::call(v, rest...);
}
};
template <typename... Functors>
bool any_call(boost::any & v, Functors const & ... f)
{
return any_call_impl<Functors...>::call(v, f...);
}
template <typename... Functors>
bool any_call(boost::any const & v, Functors const & ... f)
{
return any_call_impl<Functors...>::call(v, f...);
}
int main(void) {
boost::any a = 1;
any_call(a,
[](double d) { std::cout << "double " << d << std::endl; },
[](int i) { std::cout << "int " << i << std::endl; }
);
return 0;
}
(Demo)
这个想法是你将boost::any
或boost::any const
作为第一个参数传递给any_call
,然后传递多个lambdas。将调用其参数类型与boost::any
中包含的对象类型匹配的第一个lambda,然后any_call
将返回true。如果没有lambda匹配,any_call
将返回false。