为什么不继承C ++构造函数？ [英] Why aren't C++ constructors inherited?

问题描述

为什么在此代码中需要通过子代构造函数？我以为不会，但是当我删除它时，编译器（gcc和VS2010）抱怨。有一个优雅的解决方法吗？似乎没有必要将此垫片插入子类。

Why is the Child pass-through constructor necessary in this code? I would think that it wouldn't be, but the compiler (gcc and VS2010) complains when I remove it. Is there an elegant workaround? It just seems pointless to have to insert this shim into child classes.

class Parent
{
public:
  Parent(int i) { }
};

class Child : public Parent
{
public:
  Child(int i) : Parent(i) { }
};

int main()
{
  Child child(4);
  return 0;
}

推荐答案

因为以下内容完全有效：

Because the following is perfectly valid:

class Parent
{
public:
    Parent(int i) { }
};

class Child : public Parent
{
public:
    Child() : Parent(42) { }
};

编译器如何猜测您是否希望派生子对象具有转发的构造函数？在多重继承的情况下，两种类型的构造函数的集合可能不匹配；从基类A转发一个构造函数时，应该调用基类B上的哪个构造函数？

How is the compiler to guess whether you want the derived child to have a forwarded constructor? And in the case of multiple inheritance, the set of constructors from the two types may not match; when forwarding a constructor from base class A, which constructor on base class B should be called?

从技术上讲，我想语言设计者可能会说，如果类型具有单个基类，并且在没有显式构造函数的情况下，将转发所有父构造函数。但是，这会产生相反的问题：如果基类具有多个构造函数（包括默认构造函数），并且子级希望仅允许使用默认构造函数，则现在必须明确指定它。

Technically, I suppose the language designers could have said that if the type has a single base class and in the absence of an explicit constructor, all parent constructors are forwarded. However, that creates the opposite problem: if a base class has multiple constructors including a default constructor, and the child wishes to allow only the default constructor, this must now be explicitly specified.

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