On Mon, 26 Jan 2009 04:58:57 +1300, John Reimer <terminal.n...@gmail.com>
wrote:
Hello tim,
On Mon, 26 Jan 2009 01:14:10 +1300, Denis Koroskin <2kor...@gmail.com>
wrote:
On Sun, 25 Jan 2009 15:06:23 +0300, Tim M <a...@b.com> wrote:
On Mon, 26 Jan 2009 00:48:21 +1300, Tim M <a...@b.com> wrote:
On Mon, 26 Jan 2009 00:18:28 +1300, Denis Koroskin
<2kor...@gmail.com> wrote:
On Sun, 25 Jan 2009 08:38:18 +0300, Tim M <a...@b.com> wrote:
On Sun, 25 Jan 2009 17:56:03 +1300, John Reimer
<terminal.n...@gmail.com> wrote:
Hello tim,
On Sun, 25 Jan 2009 16:43:55 +1300, John Reimer
<terminal.n...@gmail.com> wrote:
With this code:
--------------------------------
module test5;
interface I
{
void foo();
}
class A : I {
void foo() { }
}
class B : A, I
{
alias A.foo foo;
}
void main()
{
}
--------------------------------
I get this error:
--------------------------------
class test5.B interface function I.foo is not implemented
--------------------------------
Does this make sense? I mean, shouldn't the explicit reuse of
A.foo
in B be sufficient indication to the compiler that B is
satisfying
the contract I? I'm hoping to make use of such subtleties
in
some
code, but first I have to understand the reasoning behind
this.
:)
Note that this works if I remove the interface I from B's
declaration
-- ie "class B: A" -- since, in the D language, B is not
required to
fulfull A's interface contract even though it inherits from
it.
-JJR
It look like the real bug is re-allowing B to implement
interface I
but
sometimes bug do get reported differently. Why don't you remove
I
from
B's
declaration like you said that works. It actually says here
http://www.digitalmars.com/d/1.0/interface.html "Classes cannot
derive
from an interface multiple times."
Yes, please check the link again (further down the page). D
allows you to reimplement the interface as long as class B
provides a new implementation:
"A reimplemented interface must implement all the interface
functions, it does not inherit from a super class"...
That probably could be stated a little more clearly, but that's
what it says. As for why I'm doing it, I assure you that
there's a very specific reason why I'm trying this: it is a
possible interfacing mechansim for ported software of a much
more complicated nature than this simple reduction; I reduced
it to this in order to try to understand potential iteractions
between class and interface layers. The question here was to
figure out the reasoning behind the language design, not
necessarily whether I should be doing it or not. ;-)
-JJR
This works btw:
module test;
interface I
{
void foo();
}
class A : I {
void foo() { }
}
class B : A,I
{
void foo() { A.foo(); }
}
void main()
{
}
It is too verbose and makes twice an overhead. I'd like to avoid
this
solution.
In fact, I believe that class B : A, I {} should just work.
why? I think it is perfect how it is. You can either leave A as the
class that implements I and B would implement it through
inheritance or you can to re implement I define all new
implementations and put in return super.foo(); where needed. It is
also possible to reimplement one interface function without re
implementing the whole interface.
If you are really needing to write least code you could also do
something like this but not very nice to read:
module test;
template II(char[] func)
{
const char[] II = "typeof(super." ~ func ~ "())" ~ " " ~ func ~
"() { return super." ~ func ~ "(); }" ;
}
interface I
{
void foo();
int bar();
}
class A : I
{
void foo() { }
int bar() { return 1; }
}
class B : A,I
{
//void foo() { return super.foo(); }
mixin(II!("foo"));
mixin(II!("bar"));
}
void main()
{
}
Not only I want to write less, I want my code be cleaner and run
faster.
why? I think it is perfect how it is. You can either leave A as the
class that implements I and B would implement it through inheritance
or you can to re implement I define all new implementations and put
in return super.foo(); where needed. It is also possible to
reimplement one interface function without re implementing the
whole interface.
That what /my/ solution do.
class B : A, I {}
is *absolutely* same as
class B : A, I
{
override void foo() { super.foo(); }
override int bar() { return super.bar(); }
}
Except that when you call B.foo, there is no damn double virtual
function call.
B inherits all the functions from A implicitly. You stil may override
any of the I interface functions if need be:
class B : A, I
{
override void foo() { ... }
// int bar() is inherited from A
}
Having B explicitly override all the base class virtual functions and
forward them to A implementation just to make compiler happy is
unintuitive and plain dumb to me.
C# allows that and I see absolutely no reason why D doesn't.
I think you are missing somethinghere. Change the B definition from:
class B : A, I
to just:
class B : A
then interfaces become impicit.
What do you mean? In your example above, B does not have to implement
the interface I of A. What do you mean by "interfaces become implicit"?
-JJR
I mean it will still have the foo function and you dont have to put
anything on B to make it work:
module test;
interface I
{
void foo();
}
class A : I
{
void foo() { }
}
class B : A
{
}
void main()
{
B b = new B;
b.foo(); //this works
}
