对虚拟表的困惑

Question

我在面试中被问到这个问题。下面的程序创建了多少个虚拟表，每种情况下的输出是什么。

#include <iostream> 
using namespace std;
class A
{
 virtual void func1()
 {
  cout << "0 " <<endl;
 }

 virtual void func2()
 {
  cout << "1 " <<endl;
 }
};

 class B:Public A
 {
  void func1()
  {
   cout << "2" <<endl;
  }
 };

class C:public B
{
 virtual void func2()
 {
  cout << "3" <<endl;
 }
}

int main()
{
A* objA;

B objB ;
C objC ;

//case1:
objA = &objB;
objA->func1();
//case2:
objA = &objC;
objA->func2();
objA->func1();
return 0;
}

我在B类中感到困惑，它会为B类创建vtable吗？在这种情况下，将调用哪个函数。谁能给我解释一下。

Answer 1

请参阅评论

// I'll pretend you have a #include <stdio.h> here

class A
{
 virtual void func1()
 {
  print 0; // error!!  I'll pretend this was puts("0");
 }

 virtual void func2()
 {
  print1;  // error!!  I'll pretend this was puts("1"); 
 }
};

// there is a virtual table for class A. (for two virtual methods)


 class B:Public A
 {
  void func1()
  {
   print2; // error!!  I'll pretend this was puts("2"); 
  }
 };

// there is a virtual table for class B. (for two virtual methods)


class C:public B
{
 virtual void func2()
 {
  print 3; // error!!  I'll pretend this was puts("3"); 
 }
}

// there is a virtual table for class C. (for two virtual methods)


int main()
{
A objA;
B* objB = new B();

C* objC = new C();

//case1:
objA = &objB;   // error!! left side of type A right side of type B**
objA->func1();  // error!!  operator -> on non pointer

//case2:
objA = &objC;  // error!! left side of type A right side of type B**
objA->func2(); // error!!  operator -> on non pointer
objA->func1(); // error!!  operator -> on non pointer
return 0;
}

// nothing is printed

既然您编辑了操作码，那么这里就是新版本代码的答案。请参阅评论：

#include <iostream> 
using namespace std;
class A
{
 virtual void func1()
 {
  cout << "0 " <<endl; // it's ok, but is the space supposed to be there?
 }

 virtual void func2()
 {
  cout << "1 " <<endl; // it's ok, but is the space supposed to be there?
 }
};
// there is a virtual table for class A. (for two virtual methods)

 class B:Public A  // error!! I'll pretend Public was public (lowercase)
 {
  void func1()
  {
   cout << "2" <<endl; // it's ok, but here there's no space, is that correct?
  }
 };
// there is a virtual table for class B. (for two virtual methods)

class C:public B
{
 virtual void func2()
 {
  cout << "3" <<endl; // it's ok, but here there's no space, is that correct?
 }
}
// there is a virtual table for class C. (for two virtual methods)

int main()
{
A* objA;

B objB ;
C objC ;

//case1:
objA = &objB;
objA->func1(); // outputs (to stdout) a '2' (two) and whatever a 
               // newline is on your system (could be '\r' or '\n' or both
               // or in fact anything your platform defines a newline is)
               // stdout is then flushed.
//case2:
objA = &objC;
objA->func2(); // outputs (to stdout) a '3' (three) and whatever a 
               // newline is on your system (could be '\r' or '\n' or both
               // or in fact anything your platform defines a newline is)
               // stdout is then flushed.

objA->func1(); // outputs (to stdout) a '2' (two) and whatever a 
               // newline is on your system (could be '\r' or '\n' or both
               // or in fact anything your platform defines a newline is)
               // stdout is then flushed.
return 0;
}
// the output is '2' <newline> '3' <newline> '2' <newline> 
// where the actual character(s) for <newline> are platform dependent  

Answer 2

类B的实例也是类型A的实例。您的类B调用类A的构造函数，如果B没有从A重写func1()，那么B就有一个虚拟func2()和一个来自A的func1()重写。