c++ 中的这个指针可能是 0xffffffff

Question

After read Inside the C++ Object Model , I try to call member function from vptr.

I found the following situation by accident

gcc version 9.2.0 (tdm64-1)

#include <iostream>

class Foo;
Foo* foo;

class FooBase
{
public:
    virtual void Bar1(){};
    virtual void Bar2(){};
};
class Foo : FooBase
{
public:
    void Bar1()
    {
        if (this == nullptr)
        {
            std::cout << "this == nullptr in Bar1" << std::endl;
        }
        else
        {
            std::cout << "this == " << this << " in Bar1" << std::endl;
            std::cout << "foo == " << foo << " in Bar1" << std::endl;
        }
        std::cout << "Bar1" << std::endl;
    }
};

int main()
{
    foo = new Foo;
    // std::cout << "foo == " << foo << " in main" << std::endl;

    int* vptr = reinterpret_cast<int*>(foo);
    int* bar1_ptr = reinterpret_cast<int*>(*vptr);

    typedef void (*Func1)();
    typedef void (*Func2)(Foo*);

    Func1 bar10 = (Func1)*bar1_ptr;
    Func2 bar11 = (Func2)*bar1_ptr;

    bar10(); // sometimes 'this == 0xffffffff in Bar1'
    bar11(nullptr); // this == nullptr in Bar1
    return 1;
}

When I comment std::cout << "foo == " << foo << " in main" << std::endl;

the output is

this == 0x96ae0 in Bar1
foo == 0x96ae0 in Bar1
Bar1
this == nullptr in Bar1
Bar1

But when I uncomment std::cout << "foo == " << foo << " in main" << std::endl;

the output is

foo == 0x1d6ae0 in main
this == 0xffffffff in Bar1
foo == 0x1d6ae0 in Bar1
Bar1
this == nullptr in Bar1
Bar1

Q1: Why this pointer has a default value foo == this ?

Q2: Why this pointer was changed after uncomment?

Answer 1

When you access Bar1 via Func1, this ends up being whatever was on the stack in the first arg's position. It wasnt a default value - it just happened to be there from your previous calls.

No surprise that the stack can change with change to previous calls.

As you probably found out, calling a class member functions can be implemented as a regular function that takes a class pointer as an argument. In your compiler and case, it looks as though the following are equivalent.

foo->Bar1();
bar11(foo);