c++ - 从派生类初始化 const fusion boost 列表

Question

是否可以在不使基类成为模板类的情况下将成员 fusion vector 初始化为派生类中指定的值？

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.boost.org/doc/libs/1_57_0/libs/fusion/doc/html/fusion/container/generation/functions/make_vector.html
}

我知道这行不通 - 但我希望它能让我更容易理解我的目标。

1. 延迟 vector 将包含的类型的规范，直到从它派生一个类。
2. 通过使基类成为模板类来指定 vector 应包含的类型

如果不是-最接近它的是什么？

Answer 1

不需要基类是模板的最接近的事情是使用类型删除。您可以自己滚动 ¹ 或使用 Boost Type Erasure 等。选择最适合您的。

实现它的最简单方法是boost::any:

示例

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#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;
}

打印

(test1 test)

---

¹ 例如

• On the Tension Between Object-Oriented and Generic Programming in C++
• Generating an interface without virtual functions?
• Container for boost::multi_array of same type but with different dimentionality