使函数参数默认为周围范围

Question

考虑到下面提供的以下代码/方案，是否有可能省略指定 this?

有没有办法更改代码，以便函数/构造函数自动获取周围的范围，甚至可以作为模板参数？

#include <iostream>

class AttrBase{
public:
  virtual int g()const = 0;
};

class ABase{
public:
 void reg(const char* name, AttrBase* att){
  std::cout << name << ": " << att->g()<< std::endl;
 }
};


class Attr : public AttrBase{
public: 
  Attr(const char* name, int val, ABase* parent /* = this */) // something like this
   :v(val)
  { 
    parent->reg(name, this);
  };
  int g()const override{return  v;};
  int v;
};


class D:public ABase{
  Attr a{"a", 1, this};
  Attr b{"b", 2, this};
  Attr c{"c", 3};    //< is this somehow possible
};

int main(){
 D d;
}

Answer 1

您可以添加一个成员函数来为您ABase创建Attr。

class ABase{
public:
    void reg(const char* name, AttrBase* att) {
        std::cout << name << ": " << att->g()<< std::endl;
    }

protected:
    template <typename AttrType, typename... Args>
    AttrType make(Args&&... args) {
        return AttrType{std::forward<Args>(args)..., this};
    }
};

然后我们可以将其用作

class D:public ABase {
    Attr a = make<Attr>("a", 1);
    auto b = make<Attr>("b", 2); // works with auto to avoid specifying the type twice
};