On Friday, 4 January 2019 at 21:47:59 UTC, Neia Neutuladh wrote:
On Fri, 04 Jan 2019 08:46:24 +0000, Alex wrote:
Let's assume this is right. How to force a B object to behave
like an A object? I thought casting is a possible approach...
It requires a bit of surgery:
:)
import std.stdio;
class A
{
void foo() { writeln("hello from A!"); }
}
class B : A
{
override void foo() { writeln("hello from B!"); }
}
void main()
{
auto b = new B;
auto ptrB = cast(void**)b;
ptrB[0] = A.classinfo.vtbl.ptr;
b.foo();
}
This takes advantage of the object layout used by DMD. 'vtbl'
is the virtual function table, which is basically an array of
function pointers. Each member function defined in a type (and
its super types) gets a unique index into that array.
So when you write:
b.foo();
That works out to:
(cast(void function(B))b.vtbl[5])(b);
We replace object b's vtbl with class A's, and now b is just an
A with some extra stuff allocated in its memory block.
Don't do this in production code. This is a terrible thing to
do in production code, or any code you intend to use other than
to see how D's object model works.
I see... That's cool nevertheless!
