为什么c++没有虚构造函数?
当前回答
有一个非常基本的原因:构造函数实际上是静态函数,而在c++中没有静态函数可以是虚函数。
如果你有丰富的c++经验,你就会知道静态函数和成员函数之间的区别。静态函数与CLASS相关联,而不是与对象(实例)相关联,因此它们看不到“this”指针。只有成员函数可以是虚函数,因为虚表——使“虚”工作的函数指针的隐藏表——实际上是每个对象的数据成员。
构造函数的任务是什么?它在名称中——“T”构造函数在分配T对象时初始化它们。这将自动排除它成为成员函数!对象必须先存在,然后才有this指针,从而有虚表。这意味着即使语言将构造函数视为普通函数(它并没有,由于相关原因,我不会深入讨论),它们也必须是静态成员函数。
了解这一点的一个好方法是查看“Factory”模式,特别是工厂函数。它们做你想做的事情,你会注意到,如果类T有一个工厂方法,它是ALWAYS STATIC。这是必须的。
其他回答
听可靠消息。:)
选自Bjarne Stroustrup的c++风格和技术常见问题解答为什么没有虚拟构造函数?
虚拟调用是一种在给定局部条件下完成工作的机制 信息。特别地,“virtual”允许我们调用一个函数 只知道任何接口,而不知道对象的确切类型。来 创建对象时需要完整的信息。特别是,你 需要知道你想要创建的确切类型。因此, “对构造函数的调用”不能是虚的。
FAQ条目继续给出了一种不使用虚拟构造函数实现此目的的方法的代码。
你也不应该在构造函数中调用虚函数。参见:http://www.artima.com/cppsource/nevercall.html
另外,我不确定是否真的需要虚拟构造函数。你可以在没有它的情况下实现多态构造:你可以编写一个函数,根据所需的参数构造你的对象。
A virtual-table(vtable) is made for each Class having one or more 'virtual-functions'. Whenever an Object is created of such class, it contains a 'virtual-pointer' which points to the base of corresponding vtable. Whenever there is a virtual function call, the vtable is used to resolve to the function address. Constructor can not be virtual, because when constructor of a class is executed there is no vtable in the memory, means no virtual pointer defined yet. Hence the constructor should always be non-virtual.
When a constructor is invoked, although there is no object created till that point, we still know the kind of object that is gonna be created because the specific constructor of the class to which the object belongs to has already been called. Virtual keyword associated with a function means the function of a particular object type is gonna be called. So, my thinking says that there is no need to make the virtual constructor because already the desired constructor whose object is gonna be created has been invoked and making constructor virtual is just a redundant thing to do because the object-specific constructor has already been invoked and this is same as calling class-specific function which is achieved through the virtual keyword. Although the inner implementation won’t allow virtual constructor for vptr and vtable related reasons.
Another reason is that C++ is a statically typed language and we need to know the type of a variable at compile-time. The compiler must be aware of the class type to create the object. The type of object to be created is a compile-time decision. If we make the constructor virtual then it means that we don’t need to know the type of the object at compile-time(that’s what virtual function provide. We don’t need to know the actual object and just need the base pointer to point an actual object call the pointed object’s virtual functions without knowing the type of the object) and if we don’t know the type of the object at compile time then it is against the statically typed languages. And hence, run-time polymorphism cannot be achieved. Hence, Constructor won’t be called without knowing the type of the object at compile-time. And so the idea of making a virtual constructor fails.
抛开语义上的原因不谈,在对象被构造之后才有虚表,因此虚表的指定是无用的。