为什么在c++中没有虚构造函数?

为什么c++没有虚构造函数?

当前回答

c++的虚拟构造函数是不可能的。例如，不能将构造函数标记为虚函数。试试这段代码

#include<iostream.h>
using namespace std;
class aClass
{
    public:
        virtual aClass()
        {   
        }  
};
int main()
{
    aClass a; 
}

它会导致错误。这段代码试图将构造函数声明为虚函数。现在让我们试着理解为什么要使用virtual关键字。Virtual关键字用于提供运行时多态性。例如，试试这段代码。

#include<iostream.h>
using namespace std;
class aClass
{
    public:
        aClass()
        {
            cout<<"aClass contructor\n";
        }
        ~aClass()
        {
            cout<<"aClass destructor\n";
        }

};
class anotherClass:public aClass
{

    public:
        anotherClass()
        {
            cout<<"anotherClass Constructor\n";
        }
        ~anotherClass()
        {
            cout<<"anotherClass destructor\n";
        }

};
int main()
{
    aClass* a;
    a=new anotherClass;
    delete a;   
    getchar(); 
}

In main a=new anotherClass; allocates a memory for anotherClass in a pointer a declared as type of aClass.This causes both the constructor (In aClass and anotherClass) to call automatically.So we do not need to mark constructor as virtual.Because when an object is created it must follow the chain of creation (i.e first the base and then the derived classes). But when we try to delete a delete a; it causes to call only the base destructor.So we have to handle the destructor using virtual keyword. So virtual constructor is not possible but virtual destructor is.Thanks

2013-06-23 11:47:36

其他回答

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.

2019-07-27 08:44:28

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。

2009-04-09 08:54:11

虚函数基本上提供多态行为。也就是说，当您使用的对象的动态类型与引用它的静态(编译时)类型不同时，它提供的行为适合于对象的实际类型，而不是对象的静态类型。

现在尝试将这种行为应用于构造函数。当你构造一个对象时，静态类型总是与实际的对象类型相同，因为:

要构造一个对象，构造函数需要它要创建的对象的确切类型[…]此外[…]]则不能有指向构造函数的指针

Bjarne Stroustup (P424 c++编程语言SE)

2009-04-09 08:51:52

我能想到两个原因: