HaloO,
Larry Wall wrote:
I don't think most people want to think of functions as types--it just
clutters up the type namespace.
Which contradicts their first-class status.
They can already say:
sub mysubuser ( &f where &mysub.sig ) {...}
or some such to do explicit smartmatching against the &f object.
Didn't where always require braces? And is where unary or binary?
sub foo (Int where 0..10 $x) {...}
This looks more like the juxtaposition of the two constraints
'Int' and 'where 0..10' or is where binary there? Is the above
identical to
sub foo ($x of Int where 0..10) {...} # note of
while
sub foo ($x Int where 0..10) {...}
is a syntax error, that perhaps needs
sub foo ($x where {Int 0..10}) {...}
and applys the constraint to the of type of $x. But that
contradicts your usage of where below. So this is more a
binary where that checks $x as a whole. But I have difficulties
to get a meaning from it. The only idea is that the above
fails along the lines that Int is incompatible with Scalar.
That is '$x where Foo' means that the implementation type of
$x has to be a subtype of Scalar just like '$x is Foo' demands
exactly the implementation type Foo.
Alternately if they really want the type they can say
subset mysub of Code where &mysub.sig;
sub mysubuser ( &f where mysub ) {...} # note &
sub mysubuser ( mysub $f ) {...} # note $
That is structural subtyping. The thing I wanted was nominal
subtyping of functions. In the above &f is just constraint to
the &mysub.sig which drops the reference to the name. And we
already have a syntax for that
sub foo ( &f:(Int-->Int) ) {...} # literal
sub foo ( &f:(&mysub.sig) ) {...} # per reference,
# perhaps needs flattening with |
The important point is how does one declare that a sub implements
the concept named by another sub:
sub foo {...} # think: class foo
sub bar is foo {...} # think: class bar is foo
To conflate sub and class should actually be easy because in the
background everything is an object anyway. This would e.g. allow
the has declarator in a sub:
sub mul (Int $x)
{
has Int $.factor is rw = 1;
return $x * $.factor;
}
&mul.factor = 2;
say mul(3); # prints 6
I haven't thought that through but it might even be possible to
drop & as a sigil and get it as a prefix op.
Regards, TSa.
--
"The unavoidable price of reliability is simplicity" -- C.A.R. Hoare
"Simplicity does not precede complexity, but follows it." -- A.J. Perlis
1 + 2 + 3 + 4 + ... = -1/12 -- Srinivasa Ramanujan
