Memory managment

[vavan cpp]

Hey there! Im curious a bit about placement some struct in memory.
The question is next:

Does it safe ?

struct A
{
    int a, b;
};
struct B
{
    int a;
};

void main ()
{
    A *a = new A;
    B *b = new B;

    a->a = 5;

    memcpy (b, a, sizeof B);
    cout << b->a << endl;     //5

    delete a; a = new A;

    memcpy (a, b, sizeof B);
    cout << a->a << endl;     //5
}

I would realy appreciate some useful reference.
Thx in advance.

[jumper007]

1. Why do you declare the structure B, as long as it holds only one integer ?

struct B
{
    int a;
};

2. The correct function call is sizeof(B);.

3. I guess it is safe. As long as you delete the structures after using them, there shouldn't be any problems (that means you should have a delete b; somewhere at the end).

[vavan cpp]

It`s only example, point is in that i want transform larger struct to smaler and back and worrying about lossing value of 'a'.

2. sizeof without brackets works too. At least in VS 2010.

3. Sure, and delete a; too ;)

there shouldn't be any problems

Yes, I think so too, but not assured.

[Cubbi]

Assuming you fix all compile errors and plug all memory leaks, this is indeed safe, but it's very much borderline.

The key rule here is that a pointer to the object of any standard-layout type can be reinterpret_cast to pointer to its first element (in other words, both leading padding and member reordering are prohibited in standard-layout types)

So, when memcpy() performs reinterpret_cast<unsigned char*>(b) and reinterpret_cast<unsigned char*>(a) (which is what it does before copying), it is the same as reinterpret_cast<unsigned char*>(&b->a) and reinterpret_cast<unsigned char*>(&a->a), so it will, indeed, safely copy a->a into b->a.