假设我有两个c++类:
class A
{
public:
A() { fn(); }
virtual void fn() { _n = 1; }
int getn() { return _n; }
protected:
int _n;
};
class B : public A
{
public:
B() : A() {}
virtual void fn() { _n = 2; }
};
如果我写下面的代码:
int main()
{
B b;
int n = b.getn();
}
有人可能认为n被设为2。
结果是n被设为1。为什么?
当前回答
虚表是由编译器创建的。 类对象有一个指向虚表的指针。当它开始生命时，虚表指针指向虚表 基类的。在构造函数代码的末尾，编译器生成重新指向虚表指针的代码 到类的实际虚函数表。这样可以确保调用虚函数的构造函数代码调用 这些函数的基类实现，而不是类中的重写。
其他回答
原因是c++对象的构造就像洋葱，由内而外。基类在派生类之前构造。所以，在生成B之前，必须先生成a。当调用A的构造函数时，它还不是B，因此虚函数表中仍然有A的fn()副本的条目。
Firstly,Object is created and then we assign it 's address to pointers.Constructors are called at the time of object creation and used to initializ the value of data members. Pointer to object comes into scenario after object creation. Thats why, C++ do not allows us to make constructors as virtual . .another reason is that, There is nothing like pointer to constructor , which can point to virtual constructor,because one of the property of virtual function is that it can be used by pointers only.
虚函数用于动态赋值，因为构造函数是静态的，所以我们不能将它们设为虚函数。
从构造函数或析构函数调用虚函数是危险的，应该尽可能避免。所有c++实现都应该在当前构造函数中调用在层次结构级别定义的函数的版本，而不是更进一步。
c++ FAQ Lite在第23.7节中详细介绍了这一点。我建议你阅读这篇文章(以及FAQ的其余部分)。
摘录:
[…在构造函数中，虚调用机制被禁用，因为从派生类重写还没有发生。对象是从基础开始构造的，即“先基础后派生”。 […] 销毁是“在基类之前执行派生类”，因此虚函数的行为与构造函数一样:只使用局部定义—并且不调用覆盖函数以避免触及对象的(现在已销毁的)派生类部分。
编辑修正大部分到全部(谢谢litb)
在大多数OO语言中，从构造函数调用多态函数是导致灾难的原因。遇到这种情况时，不同的语言会有不同的表现。
基本问题是，在所有语言中，基类型必须在派生类型之前构造。现在，问题是从构造函数调用多态方法意味着什么。你希望它表现得怎样?有两种方法:在基本层调用方法(c++风格)或在层次结构底部的未构造对象上调用多态方法(Java方式)。
In C++ the Base class will build its version of the virtual method table prior to entering its own construction. At this point a call to the virtual method will end up calling the Base version of the method or producing a pure virtual method called in case it has no implementation at that level of the hierarchy. After the Base has been fully constructed, the compiler will start building the Derived class, and it will override the method pointers to point to the implementations in the next level of the hierarchy.
class Base {
public:
Base() { f(); }
virtual void f() { std::cout << "Base" << std::endl; }
};
class Derived : public Base
{
public:
Derived() : Base() {}
virtual void f() { std::cout << "Derived" << std::endl; }
};
int main() {
Derived d;
}
// outputs: "Base" as the vtable still points to Base::f() when Base::Base() is run
In Java, the compiler will build the virtual table equivalent at the very first step of construction, prior to entering the Base constructor or Derived constructor. The implications are different (and to my likings more dangerous). If the base class constructor calls a method that is overriden in the derived class the call will actually be handled at the derived level calling a method on an unconstructed object, yielding unexpected results. All attributes of the derived class that are initialized inside the constructor block are yet uninitialized, including 'final' attributes. Elements that have a default value defined at the class level will have that value.
public class Base {
public Base() { polymorphic(); }
public void polymorphic() {
System.out.println( "Base" );
}
}
public class Derived extends Base
{
final int x;
public Derived( int value ) {
x = value;
polymorphic();
}
public void polymorphic() {
System.out.println( "Derived: " + x );
}
public static void main( String args[] ) {
Derived d = new Derived( 5 );
}
}
// outputs: Derived 0
// Derived 5
// ... so much for final attributes never changing :P
如您所见，调用多态(c++术语为虚拟)方法是一个常见的错误来源。在c++中，至少你可以保证它永远不会对一个尚未构造的对象调用方法……
c++ FAQ Lite很好地涵盖了这一点:
本质上，在调用基类构造函数期间，对象还不是派生类型，因此调用的是基类型的虚函数实现，而不是派生类型的实现。
