如何在C ++ 03中伪造构造函数继承？ [英] How can I fake constructor inheritance in C++03?

问题描述

据我所知，你不能继承C ++中的构造函数。但是在某些情况下，它可能需要看起来像你可以实例化继承的类，就像你实例化它们的基地一样：

  struct Base {
 int i; 
 int const j; 
 Base（int i，int j）：i（i），j（j）{} 
}; 
 
 //选项1）
 struct Derived1：Base {
 Derived1（int i，int j）：Base（i，j）{} 
} 
 Base * baseFactory（）{
 return new Derived1（42,77）; 
} 
 
 //选项2）
 struct Derived2：Base {
}; 
 Base * baseFactory（）{
 Base * b = new Derived2（）; 
 b-> i = 42; 
 b-> j = 47; //破了！ 
 //替代方法可以实现Base :: operator = 
 return b; 
} 
  

请注意，派生类可以是默认构造的，基本类。

选项1是通常做的，我想，但你是打字的代码，没有表示任何新的。选项2打破 const 正确性（并防止您使用引用作为成员）。因为一切都必须是可分配的。

编辑：C ++ 11是伟大的，但不幸的是，我需要一个C ++ 03解决方案。

/ p>

解决方案

有一种替代方法：

  // option 3）
 struct Derived3：Base {
 Derived3（const Base& b）：Base（b）{} 
}; 
 Base * baseFactory3（）{
 return new Derived3（Base（42,47））; 
} 
  

这可能不是一个好主意，如果构建一个完整的Base对象是昂贵的，需要外部资源。在这种情况下，您可以创建一个轻量级对象，其中携带Base的构造函数参数，因此，您可以使用多个Base构造函数，而不是多个BaseArguments构造函数，以及一个包含BaseArguments的Base构造函数。然而，我认为在大多数情况下，很多人不会考虑这种良好的风格。

As far as I know, you cannot inherit constructors in C++. But there are situations, where it might be required that it looks like you can instantiate inherited classes the same way you instantiate their base:

struct Base {
  int i;
  int const j;
  Base(int i, int j) : i(i), j(j) {}
};

// option 1)
struct Derived1 : Base {
  Derived1(int i, int j) : Base(i,j) {}
};
Base* baseFactory() {
  return new Derived1(42,47);
}

// option 2)
struct Derived2 : Base {
};
Base* baseFactory() {
  Base* b = new Derived2();
  b->i = 42;
  b->j = 47; // broken!
  // alternative could be to implement Base::operator=
  return b;
}

Note that the derived classes could be default constructed if it weren't for their base class(es).

Option 1 is what is usually done, I think, but you are typing code without expressing anything new. Option 2 breaks const correctness (and prevents you from using references as members) because everything must be assignable.

Is there a nice way around this?

Edit: C++11 is great, but unfortunately I need a C++03 solution.

解决方案

There is an alternative approach:

// option 3)
struct Derived3 : Base {
  Derived3(const Base &b) : Base(b) {}
};
Base* baseFactory3() {
    return new Derived3(Base(42,47));
}

This may not be a good idea if constructing a full Base object is expensive or requires external resources. In that case you could create a light-weight object that carries constructor arguments for Base, so that instead of multiple Base constructors you have multiple BaseArguments constructors, and one Base constructor that takes in BaseArguments. However I don't think many people would consider that good style in most circumstances.

这篇关于如何在C ++ 03中伪造构造函数继承？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！