I feel that c ++ is a bad place

aleckray

#include <iostream.h>

class A
{
public:
int v;
A () {}
virtual void f () {cout << "A :: f ()" << endl;}
virtual ~ A () {}
};

class B: public A
{
public:
B () {}
void f () {cout << "B :: f ()" << endl;}
~ B () {}
};

int main (int argc, char * argv [])
{
A * pa = new B ();
pa-> v = 1;
A c;
c.v = 2;

A&ra = * pa;
ra.f ();
ra = c;
ra.f ();

delete pa;
return 0;
}

The above program will output two lines of B :: f ()

Why not copy the virtual table too? If you don't know it, it is often debugged for a long time because of this mechanism.

g791113

This has nothing to do with references. In C ++, if the object is copied by byte, the virtual function table pointer will not be copied in together. Unless hard to come: unsigned char * p = (unsigned char *) pa;
    unsigned char * q = (unsigned char *)&c;
for (int i = 0; i <sizeof (c); i ++)
    p [i] = q [i];

珈蓝易拉罐

Why not copy the virtual table too?
Since it is "copy", not just "pointing" or "referencing", what polymorphism is there?
You can't copy it as an object of class X and call a function of class Y at the same time.

g791113

At first glance, the landlord has never studied software engineering.

珈蓝易拉罐

Oh, I didn't think about the landlord's problem clearly, and I talked about it.
The first floor is right, "object copy" does not affect virtual tables.
Ra was originally referencing a class B object, and this would not change.
ra = c;
This sentence just modifies the base class sub-object of that referenced object. You cannot modify the virtual table. Once you modify it, it means that it still refers to a class b object (), but it has a virtual table of class A. Then, the following sentence:
delete pa;
It will not work, and the destructor will not be called accurately. When the class is complicated, it is easy to get a memory leak.

g791113

#include <stdlib.h>

int main (int argc, char * argv [])
{
A * pa = new B ();
pa-> v = 1;
A c;
c.v = 2;

A&ra = * pa;
ra.f ();

/ *
unsigned char * p = (unsigned char *) pa;
    unsigned char * q = (unsigned char *)&c;

for (int i = 0; i <sizeof (c); i ++)
p [i] = q [i];
* /
memcpy (pa,&c, sizeof (c));

ra.f ();

delete pa;
return 0;
}

I think this will probably please the landlord.

It really doesn't make any sense except for occasional entertainment. The class B object that was originally allocated is later copied into the class A object member. But this is also normal and can be used as some special initialization.
But later it is wrong, copy the virtual function table pointer into it, and the result is that the original class B object is converted to a class A object at runtime (a veritable runtime type conversion -_-!). So does ra refer to a class A object or a class B object?

A a;

B b;

The true meaning of b = a; is: copy the parent domain members of the parent object a to the child B object in bytes (assuming that the equal sign is not overloaded). If you also copy the virtual function pointer to b, it is equivalent to losing your qualification as class B, that is, assimilating with A.

Moderator Morning Star is right.

Chinawinner

The semantics of assignments are relative to objects, while virtual function tables are relative to classes. Since the assignment only operates on the object, of course, the type conversion of the assigned object is not automatically performed (the compiler does not know the actual type of the referenced object during the assignment process). The type of a reference implies that the user manipulates the abstract form of the referenced object, but it does not (and should not) intervene in changing the type of the actual object.

sinsun123

Ha ha, a topic worth thinking about, collect first!

zhilong821

The landlord is still not familiar with the mechanism of polymorphism.

finefineboy

Strange, this program does not report warnning even if you add the -Wall parameter on g ++ ???????

<< Thinking in C ++ >> vol.1 charpter 11

Once a reference is initialized to an object, it cannot be
changed to refer to another object. (Pointers can be pointed to
another object at any time.)

<< ISO C ++ 14882-2003 >> 8.5.3.2
A reference cannot be changed to refer to another object after initialization.