Re: Constructing Derived in shell of Base <shudder>
* cpp4ever, on 18.07.2010 18:02:
On 07/15/2010 09:29 PM, Pete Becker wrote:
On 2010-07-15 16:17:44 -0400, Kai-Uwe Bux said:
cpp4ever wrote:
[...]
may I suggest the following
//---------------------
struct B { int j; virtual void f() { std::cout<< "B::f\n"; } virtual
~B() {} };
struct D : B { virtual void f() { std::cout<< "D::f\n";} };
int main()
{
char obj[sizeof(D)];
B *p = new (obj) B;
p->f(); // B::f() involked
p->~B();
B *p = new (obj) D;
Probably, the above was meant to be:
p = new (obj) D;
Nice idea: assigning a new value to p circumvents [3.8/7] as no old
pointer
to corps is in continued use.
Well, sure. That's the essence of what every memory manager does.
p->f(); // D::f() involked
p->~B(); // D Destructor will be called
}
//---------------------
This still requires care, as before creating an object with new, any
previously created object should be destroyed. Worse still the object
will not automatically be destroyed when it goes out of scope. Therefore
IMHO this sort of functionality should be encapsulated within a class
that ensures things are handled safely.
Just a question: does the standard guarantee that obj is correctly
alligned
for objects of type D and B?
No. It's a char array, with alignment suitable for a char array.
In C++0x you'll get aligned storage with the aligned_storage template:
aligned_storage<sizeof(D), alignof(D)>::type obj;
or with attributes:
unsigned char obj [[ align(alignof(D)) ]] [sizeof(D)];
Does alignment really apply to where a variable is stored?
Yes.
The basic thing to understand about alignment is what it is. For that you have
to think of the hardware level, and imagine an old-fashioned parallel address
bus, one with one line per address bit, and a ditto parallel data bus. Comes a
write signal on the control bus, and the electronics has to connect the data bus
wires to the right memory circuits, depending on the address on the address bus.
Well it gets complicated if say a 32 bit data bus should be able to connect its
lowest byte to any byte whatsoever. Instead for a 32-bit write signal it can
connect the lowest byte on the bus only to every fourth memory byte, say. Then
you have alignment 4, from /the start of memory/, address 0. Anything else may
cause a trap or be sloooow. In one Windows version you could tell Windows to
translate the trap into sloooow byte-by-byte ops.
Cheers & hth.,
- Alf
--
blog at <url: http://alfps.wordpress.com>