BUILD and other submethods

[Ingo Blechschmidt]

Hi,

class Foo {
submethod BUILD() {
say 42;
}
}

class Bar is Foo {
submethod BUILD() {
say 23;
}
}

my Bar $bar .= new;

I suppose this will output:
42
23

S12 says that "submethod[s] [are] called only when a method call is
dispatched directly to the current class", but the default implementation of
"new" (as given by Object), calls BUILDALL, which in turn calls all BUILDs, so
I think both BUILDs are executed. Are they?

S12 says that "[s]ubmethods are for declaring infrastructural methods that
shouldn't be inherited by subclasses". I read this "shouldn't be inherited" as
that a submethod definition of the same name in a subclass does not
overwrite/substitute the definition of the parent class. Correct?


class A { submethod blarb() { say 42 } }
class B is A { submethod blarb() { say 23 } }
B.new.blarb;

Does this only output "23"? (I think so, as I don't call A::blarb explicitly.
If I wanted to call A's blarb, I'd have to say "B.new.A::blarb", correct?)


--Ingo
</lots_of_questions>

--
Linux, the choice of a GNU | self-reference, n. - See self-reference
generation on a dual AMD |
Athlon! |

[Larry Wall]

On Mon, May 16, 2005 at 02:59:11PM +0000, Ingo Blechschmidt wrote:
: Hi,

:
: class Foo {
: submethod BUILD() {
: say 42;
: }
: }
:
: class Bar is Foo {
: submethod BUILD() {
: say 23;
: }
: }
:
: my Bar $bar .= new;
:
: I suppose this will output:
: 42
: 23
:
: S12 says that "submethod[s] [are] called only when a method call is
: dispatched directly to the current class", but the default implementation of
: "new" (as given by Object), calls BUILDALL, which in turn calls all BUILDs, so
: I think both BUILDs are executed. Are they?

Yes.

: S12 says that "[s]ubmethods are for declaring infrastructural methods that

: shouldn't be inherited by subclasses". I read this "shouldn't be inherited" as
: that a submethod definition of the same name in a subclass does not
: overwrite/substitute the definition of the parent class. Correct?

Correct.

: class A { submethod blarb() { say 42 } }

: class B is A { submethod blarb() { say 23 } }
: B.new.blarb;
:
: Does this only output "23"? (I think so, as I don't call A::blarb explicitly.
: If I wanted to call A's blarb, I'd have to say "B.new.A::blarb", correct?)

Yes.

Larry