C++ 从派生类初始化const fusion boost列表
是否可以将成员融合向量初始化为派生类中指定的值,而不使基类成为模板类 像这样:C++ 从派生类初始化const fusion boost列表,c++,boost,c++14,fusion,C++,Boost,C++14,Fusion,是否可以将成员融合向量初始化为派生类中指定的值,而不使基类成为模板类 像这样: class container { const auto children; container (auto children):children (children){} } class derived : public container { derived():container(make_vector(string("test1"),string("test"))){} // http://www.boos
class container
{
const auto children;
container (auto children):children (children){}
}
class derived : public container
{
derived():container(make_vector(string("test1"),string("test"))){} // http://www.boost.org/doc/libs/1_57_0/libs/fusion/doc/html/fusion/container/generation/functions/make_vector.html
}
我知道这是行不通的,但我希望这能让我更容易理解我的目标
如果不是,那么最接近它的是什么?不要求基类成为模板的最接近的事情是使用类型擦除。您可以使用自己的ª或使用Boost类型擦除等。选择最适合您的 实现它的最简单方法是
boost::any
:
样品
#include <boost/any.hpp>
#include <boost/fusion/include/io.hpp>
#include <boost/fusion/include/vector.hpp>
#include <boost/fusion/include/make_vector.hpp>
#include <string>
namespace fus = boost::fusion;
class container
{
protected:
boost::any children;
template <typename T>
container (T const& children) : children(children) {}
};
class derived : public container
{
using V = boost::fusion::vector2<std::string, std::string>;
public:
derived() :
container(fus::make_vector(std::string("test1"),std::string("test"))){}
friend std::ostream& operator<<(std::ostream& os, derived const& d) {
return os << boost::any_cast<V const&>(d.children);
}
};
#include <iostream>
int main() {
derived d;
std::cout << d;
}
例如
(test1 test)