在C ++中从基础调用潜在孩子的构造函数 [英] Calling constructor of potential child from base in C++

问题描述

如何实现这种模式：

  class Base {//doesn't know anything about potential descendant-classes, like Child
  public:
      Base * foo( void) {
        //some code
          return ( Base *) new !Child-constructor!();
      }
  };

  class Child : public Base { };

//—————————————————————————————————————————————————
  #include <iostream>
  #include <typeinfo>

  using namespace std;

  int main( void) {
      Base * base_p = Child().f();

      cout << typeid( *base_p).name(); //expected to get "Child"

  return 0;
  }

我找不到这种类型的结构的正确语法潜在子构造函数）。

I can't find the right syntax for this type of constructions (calling the "potential" child constructor).

UPD ：我忘了提及（没有认为可以有误解） c $ c> Child 类在 Base 的定义中不可知。因此，我想要 foo 来调用要继承的类的构造函数。

UPD: I forgot to mention (didn't think that there can be missunderstanding), that Child class must not be known in definition of Base. So I wanted foo to call the constructor of the class in which it's gonna be inherited.

推荐答案

...在Base的定义中不能知道Child类，所以我想要foo来调用类继承的类的构造函数。

"... that Child class must not be known in definition of Base. So I wanted foo to call the constructor of the class in which it's gonna be inherited."

IMHO最简单的方法是提供 模板工厂功能 c

IMHO the easiest way is to provide a templated factory function with Base

class Base {
public:
    template<class Derived>
    static std::unique_ptr<Base> foo( void) {
      //some code
      return std::unique_ptr<Base>(new Derived());
    }
};

class Child : public Base {
public:
    Child() {}
    virtual ~Child() {}
};

int main() {
    std::unique_ptr<Base> p = Base::foo<Child>();
    return 0;
}

检查可编译的样本这里请。

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