Vector changing the size of class - c++

Question

EDIT:
Though badly formatted this Question had a nice catch.So, I am editing this to retain this in a better format for future visitors who stumble across this question.
In the code sample below can someone please explain Why is the size of class different than expected after memcpy? What is the reason?
Here is the online demo on Ideone.
#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
class A
{
public:
int a;
virtual void f1() { cout <<"inside a::f1\n"; }
A() { a = 1; }
};
class B
{
public:
int b;
virtual void f2() { cout <<"inside b::f2\n"; }
virtual void f5() { cout <<"inside b::f5\n"; }
B() { b = 2; }
};
class C : public A, public B
{
public:
int c;
void f1() { cout <<"inside c::f1\n"; }
void f2() { cout <<"inside c::f2\n"; }
virtual void f3() { cout <<"inside c::f3\n"; }
virtual void f4() { cout <<"inside c::f4\n"; }
C() { c = 3; }
};
int fun()
{
int a = 1;
return a * 2;
}
int main()
{
C c;
C c2;
int (*g)() = &fun;
void (A::*g1)() = &A::f1;
void (C::*g2)();
g2 = &C::f1;
(c.*g2)();
printf("%p\n",g2);
cout << sizeof(g2) << endl;
g2 = &C::f2;
(c.*g2)();
printf("%p\n", g2);
// Why is the output 1 here in g++ or visual C++?
cout << g2;
// Why is the sizeof returning 8? Please explain.
cout << sizeof(g2) << endl;
g2 = &C::f1;
std::vector<unsigned char> a_vec(sizeof(g2));
memcpy(&a_vec[0], &g2, sizeof(g2));
for(size_t i = 0; i < sizeof(g2); ++i)
{
cout << hex << static_cast<unsigned>(a_vec[i]) << " ";
}
cout << endl;
g2 = &C::f2;
std::vector<unsigned char> a_vec1(sizeof(g2));
memcpy(&a_vec1[0], &g2, sizeof(g2));
for(size_t i = 0; i < sizeof(g2); ++i)
{
cout << hex << static_cast<unsigned>(a_vec1[i]) << " ";
}
cout << endl;
cout << sizeof(g) <<endl;
cout << sizeof(g1) <<endl;
cout << sizeof(g2) <<endl;
// Why is sizeof(C) printing 14 and not 20 in visual C++?
// If yes why is this so ?
cout << sizeof(C) << endl;
cout << sizeof(c2) << endl;
cout << (&c) << endl;
cout << c.a << endl;
cout << c.b << endl;
cout << c.c << endl;
return 0;
}
From the above code sample the Output I get is:
inside c::f1
0x1
8
inside c::f2
0x5
18
1 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0
4
8
8
14
14
0xbffe375c
1
2
3
Following are my Questions:
Why is the output 1 here in g++ or visual C++?
cout << g2;
Why is the sizeof returning 8? Please explain.
cout << sizeof(g2) << endl;
Why is sizeof(C) printing 14 and not 20 in visual C++? If yes why is this so?
cout << sizeof(C) << endl;

Answer 1

Why is the sizeof returning 8. Please explain?
cout <<sizeof(g2)<<endl;
returns 8 because g2 is a pointer and size of an pointer on your enviornment is 8.
Why is the output 1 here in g++ or visual C++?
cout << g2;
The << operator does not have an overloaded version which accepts a pointer. So the pointer gets converted to a bool type, with a value 1, and cout prints it.
The C++ Standard allows this:
C++03 Standard 4.12 Boolean conversions
1 An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool.
Why is sizeof(C) printing 14 and not 20 in visual C++.
cout<<sizeof(C)<<endl;
It displays the size of C correctly just in hexadecimal(14 in hex == 20 in decimal). This is because you used the hex I/O manippulator to print an address before.
cout<<dec<<sizeof(C)<<endl;
will set the I/O manipulator to decimal mode again and it will output 20 as you expect.
A word about printf and type safety:
printf is not type safe.When using printf it is the users responsibility to pass the proper formart descriptor and data type to it. If there is a mismatch then an Undefined Behavior will occur. An Undefined behavior means that the program might crash or show any weird behavior.
printf( "%p\n", g2 );
Is an example of Undefined Behavior, there is a mismatch in the format descriptor and the data type. Note that the compiler does complain about this and you should always look out and check such warnings emitted by the compiler.
warning: format ‘%p’ expects type ‘void*’, but argument 2 has type ‘void (C::*)()’

Answer 2

The results of sizeof for a given type never change. (At least for a
conforming C++ program. g++, I believe, supports VLAs in C++, as an
extension, and the sizeof an object containing a VLA might change.)
With regards to your explicit questions:
printf( "%p\n", g2 );
is undefined behavior. You're passing a pointer to member to an output
formatter which requires a void*. (g++ will warn about this, I
believe, at least with certain options.) You might get just about
anything (including a program crash). Pointers to members are not
void*, and can't be converted to void*.
cout << g2;
I can't believe that this compiles.
cout << sizeof(g2) << endl;
sizeof(g2) returns 8 because this is the size of a pointer to member
on your system.
cout << sizeof(C) << endl;
Prints 14 because this is the size in hexadecimal of an object of type
C on your system. In other words, sizeof(C) is 20. Since you've
set output for hex, and never reset it, the output is hexadecimal.