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。

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.

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。

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。

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!

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)