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

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。

其他回答

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。

为了回答当你运行这段代码时会发生什么/为什么,我通过编译它 g++ -ggdb main。Cc,并逐步使用gdb。

main.cc:

class A { 
  public:
    A() {
      fn();
    }
    virtual void fn() { _n=1; }
    int getn() { return _n; }

  protected:
    int _n;
};


class B: public A {
  public:
    B() {
      // fn();
    }
    void fn() override {
      _n = 2;
    }
};


int main() {
  B b;
}

在main处设置断点,然后进入B(),打印this ptr,进入a()(基构造函数):

(gdb) step
B::B (this=0x7fffffffde80) at main2.cc:16
16    B() {
(gdb) p this
$27 = (B * const) 0x7fffffffde80
(gdb) p *this
$28 = {<A> = {_vptr.A = 0x7fffffffdf80, _n = 0}, <No data fields>}
(gdb) s
A::A (this=0x7fffffffde80) at main2.cc:3
3     A() {
(gdb) p this
$29 = (A * const) 0x7fffffffde80

显示它最初指向派生的B对象B,该对象B被构造在0x7fffffffde80的堆栈上。下一步是进入以A()为基数的ctor,这变成了A * const到相同的地址,这是有意义的,因为以A为基数的对象正好在B对象的开头。但它仍然没有被构建:

(gdb) p *this
$30 = {_vptr.A = 0x7fffffffdf80, _n = 0}

还有一步:

(gdb) s
4       fn();
(gdb) p *this
$31 = {_vptr.A = 0x402038 <vtable for A+16>, _n = 0}

_n已经初始化,它的虚函数表指针包含虚void A::fn()的地址:

(gdb) p fn
$32 = {void (A * const)} 0x40114a <A::fn()>
(gdb) x/1a 0x402038
0x402038 <_ZTV1A+16>:   0x40114a <_ZN1A2fnEv>

因此,下一步通过this->fn()执行A::fn()是完全有意义的,前提是活动this和_vptr.A。再一步,我们回到B() ctor:

(gdb) s
B::B (this=0x7fffffffde80) at main2.cc:18
18    }
(gdb) p this
$34 = (B * const) 0x7fffffffde80
(gdb) p *this
$35 = {<A> = {_vptr.A = 0x402020 <vtable for B+16>, _n = 1}, <No data     fields>}

碱基A已经构造好了。注意,存储在虚函数表指针中的地址已经更改为派生类B的虚表,因此调用fn()将通过this->fn()选择派生类重写B::fn(),给定活动this和_vptr。A(在B()中调用B::fn()来查看这个。)再次检查存储在_vptr中的1个地址。A显示它现在指向派生类重写:

(gdb) p fn
$36 = {void (B * const)} 0x401188 <B::fn()>
(gdb) x/1a 0x402020
0x402020 <_ZTV1B+16>:   0x401188 <_ZN1B2fnEv>

By looking at this example, and by looking at one with a 3 level inheritance, it appears that as the compiler descends to construct the base sub-objects, the type of this* and the corresponding address in _vptr.A change to reflect the current sub-object being constructed, - so it gets left pointing to the most derived type's. So we would expect virtual functions called from within ctors to choose the function for that level, i.e., same result as if they were non-virtual.. Likewise for dtors but in reverse. And this becomes a ptr to member while members are being constructed so they also properly call any virtual functions that are defined for them.

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

解决这个问题的一个方法是使用工厂方法来创建对象。

为你的类层次结构定义一个公共基类,其中包含一个虚拟方法afterConstruction():

class Object
{
public:
  virtual void afterConstruction() {}
  // ...
};

定义一个工厂方法:

template< class C >
C* factoryNew()
{
  C* pObject = new C();
  pObject->afterConstruction();

  return pObject;
}

像这样使用它:

class MyClass : public Object 
{
public:
  virtual void afterConstruction()
  {
    // do something.
  }
  // ...
};

MyClass* pMyObject = factoryNew();

我刚刚在一个程序中出现了这个错误。 我有这样的想法:如果方法在构造函数中被标记为纯虚函数会发生什么?

class Base {
public:
    virtual int getInt() = 0;
    
    Base(){
        printf("int=%d\n", getInt());
    }
};

class Derived : public Base {
    public:
        virtual int getInt() override {return 1;}
};

和…有趣的事情!你首先得到编译器的警告:

warning: pure virtual ‘virtual int Base::getInt() const’ called from constructor

和一个来自ld的错误!

/usr/bin/ld: /tmp/ccsaJnuH.o: in function `Base::Base()':
main.cpp:(.text._ZN4BaseC2Ev[_ZN4BaseC5Ev]+0x26): undefined reference to `Base::getInt()'
collect2: error: ld returned 1 exit status

这是完全不合逻辑的,你只得到一个警告从编译器!