[perl #61130] :nth() does not work with :x() or :g in .subst in Rakudo

[Moritz Lenz via RT]

On Sun Dec 07 07:24:07 2008, masak wrote:
> The .subst method in Rakudo r33599 can understand :x()...
> 
> $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :x(2))' # yes
> bar1bar2foo3foo4
> 
> ...and :nth()...
> 
> $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :nth(2))' # yes
> foo1bar2foo3foo4
> 
> ...and :g...
> 
> $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :g)' # yes
> bar1bar2bar3bar4
> 
> ...but not :x() together with :nth()...
> 
> $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :x(2),
> :nth(2))' # expected foo1bar2foo3bar4
> foo1bar2foo3foo4
> 
> ...and not :g together with :nth().
> 
> $ perl6 -e 'say "foo1foo2foo3foo4".subst("foo", "bar", :g, :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)