On Sun, Dec 07, 2008 at 03:09:30PM -0800, Moritz Lenz via RT wrote:
> > ...but not :x() together with :nth()...
> >
> > $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :x(2),
> > :nth(2))' # expected foo1bar2foo3bar4
> > foo1bar2foo3foo4
> >
> > The above are my personal expectations. The current version of S05 is
> > silent on how :nth() interacts with :x() and :g. There are spectests
> > for :g:nth but not (as far as I can see) for :x:nth.
>
> Since your personal expectations are the same as mine, I took the liberty to
> turn our expectations into spec tests, in
> t/spec/S05-substitution/subst.t (pugs r24207).
>
> The reasoning behind it is quite simple: I imagine :g to mean the same as
> :x(*). Now a :x($x) and :nth($n) interact like this:
>
> for 1 .. $x {
> match here
> if ($x-1) % $n == 0 {
> do substitution
> }
> }
>
> (CC'ing p6l, since it defines language semantics, albeit just a bit)
The problem with this reasoning is that :nth doesn't have to be an
integer -- S05 also allows things like
:nth(1,2,3,5,8,13)
:nth({.is_fibonacci})
It's not immediately obvious how those forms of :nth would interact
with :x or :g.
The way to do something like "replace every 3rd occurrence" would
seem to be to do things like:
:nth({ $_ % 3 == 0 })
:nth(3..*:by(3))
Another problem is that :x($n) specifies "perform $n substitutions",
such that if we say :x(4) and there aren't four things to substitute,
then none of the substitutions take place.
At the moment I think that :x specifies a constraint on the
total number of substitutions that are (must be) performed,
while :nth selects the candidate matches to be substituted.
Thus
.subst( $pat, $rep, :nth(1,4,9,16,25,36), :x(4) )
would perform substitutions on the 1st, 4th, 9th, and 16th matches,
and do nothing if at least sixteen matches were not found.
Pm
