On Mon, 08 Feb 2010 17:19:28 -0500, Andrei Alexandrescu
<[email protected]> wrote:
Steven Schveighoffer wrote:
On Mon, 08 Feb 2010 16:09:19 -0500, Andrei Alexandrescu
<[email protected]> wrote:
Steven Schveighoffer wrote:
On Mon, 08 Feb 2010 15:13:33 -0500, Andrei Alexandrescu
<[email protected]> wrote:
Steven Schveighoffer wrote:
On Mon, 08 Feb 2010 14:36:37 -0500, Andrei Alexandrescu
<[email protected]> wrote:
Don wrote:
I don't understand this. How does belowTop() know how to call
top()?
It calls top() through the normal interface mechanism. Inside
belowTop(), this has Stack!T type.
Actually, I think Don has a point here. A virtual function (even
on an interface) is always called with the 'this' pointer, not the
interface pointer.
That is done via an adjustment of the reference. In the case of an
interface method, no adjustment is necessary. Inside the method,
"this" has the static type of the interface and the dynamic type
whichever class implements the interface.
void foo(Stack!T st)
{
auto x = st.belowTop();
}
OK, so if st's virtual function for belowTop points to the default
implementation, there is no adjustment. But what if the actual
object *did* override the default implementation? Does it also
receive the interface pointer as 'this'? Where does the adjustment
happen? What happens if you have a reference to the actual concrete
object type? Do you have to thunk to the correct interface to pass
in the expected interface pointer? It can't be both ways.
If an object overrode the default implementation, the pointer to
method belowTop will point to code that does do the adjustment.
If I understand this correctly, calling such a "default
implementation" function is different than calling a standard interface
function. And each entry in such an interface for a overridden method
will point to a "pre function" that adjusts the 'this' reference before
jumping to the real implementation.
Actually that's what's happening today.
Yes, but I think it's happening at the call site and not inside the
function itself.
I will run a test...
You are right, I was wrong. So this is already the way it works (good to
know!):
interface I
{
void foo();
}
interface J
{
void foo();
}
class C : I, J
{
int i;
void foo() { i = 5;}
}
produces the following asm for C.foo:
_D13testinterface1C3fooMFZv:
push EBP
mov EBP,ESP
sub ESP,4
mov dword ptr 8[EAX],5
leave
ret
nop
_TMP0:
add EAX,0FFFFFFF4h
jmp near ptr _D13testinterface1C3fooMFZv
_TMP1:
add EAX,0FFFFFFF0h
jmp near ptr _D13testinterface1C3fooMFZv
where _TMP0 and _TMP1 are the little pre functions that get stored in the
interface vtables.
I therefore don't think there are any issues, I misunderstood the way
interface functions work. I didn't realize the interface function called
a little pre function. I thought it was done at the call site before the
call.
The vtable entries of the object itself would point to a function that
does not do the adjustment, correct?
Yes, but let's not forget that each object stores more than one vptrs.
Right, I just thought the different vtables contained identical references
to the same functions. It makes sense that is not the case.
I think all the information is available to make this work, the only
issue I see is that a function with a default implementation
artificially changes the ABI for that function. Adding a default
implementation therefore will make compiled objects incompatible, even
with the same vtable layout.
Not getting this, but I'll let Walter weigh in.
It's my bad, I thought normal interface calls were different than this
method of having a little "pre" function, but it makes sense to do it that
way.
Sorry for the noise. I retract my objection on these grounds :)
-Steve
