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

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)

其他回答

作为补充,调用尚未完成构造的对象的虚函数也将面临同样的问题。

例如,在对象的构造函数中启动一个新线程,并将该对象传递给新线程,如果新线程在对象完成构造之前调用该对象的虚函数将导致意想不到的结果。

例如:

#include <thread>
#include <string>
#include <iostream>
#include <chrono>

class Base
{
public:
  Base()
  {
    std::thread worker([this] {
      // This will print "Base" rather than "Sub".
      this->Print();
    });
    worker.detach();
    // Try comment out this code to see different output.
    std::this_thread::sleep_for(std::chrono::seconds(1));
  }
  virtual void Print()
  {
    std::cout << "Base" << std::endl;
  }
};

class Sub : public Base
{
public:
  void Print() override
  {
    std::cout << "Sub" << std::endl;
  }
};

int main()
{
  Sub sub;
  sub.Print();
  getchar();
  return 0;
}

这将输出:

Base
Sub

在对象的构造函数调用期间,虚函数指针表没有完全构建。这样做通常不会给你带来你期望的行为。在这种情况下调用虚函数可能有效,但不能保证,应该避免使用,以便便于移植并遵循c++标准。

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……

你知道Windows资源管理器的崩溃错误吗?“纯虚函数调用…” 同样的问题…

class AbstractClass 
{
public:
    AbstractClass( ){
        //if you call pureVitualFunction I will crash...
    }
    virtual void pureVitualFunction() = 0;
};

由于pureVitualFunction()函数没有实现,并且在构造函数中调用该函数,因此程序将崩溃。

我看不出这里虚拟关键词的重要性。B是一个静态类型变量,它的类型由编译器在编译时确定。函数调用不会引用虚表。当b被构造时,它的父类的构造函数被调用,这就是为什么_n的值被设置为1。