I've been thinking about this issue some more and it occurs to me that we
might be thinking about this the wrong way.
Providing a :fillin() adverb on C<zip> is a suboptimal solution, because it
implies that you would always want to fill in *any* gap with the same value.
While that's likely in a two-way zip, it seems much less likely in a multiway zip.
So I now propose that C<zip> works like this:
C<zip> interleaves elements from each of its arguments until
any argument is (a) exhausted of elements I<and> (b) doesn't have
a C<fill> property.
Once C<zip> stops zipping, if any other element has a known finite
number of unexhausted elements remaining, the <zip> fails.
In other words, you get:
@i3 = 1..3 ;
@i4 = 1..4 ;
@a3 = 'a'..'c' ;
zip(@a3, @i3) # 'a', 1, 'b', 2, 'c', 3
zip(@i3, @i4) # fail
zip(100..., @a3, @i3) # 100, 'a', 1, 101, 'b', 2, 102, 'c', 3
zip(100..., @a3, @i4) # fail
zip(@a3 but fill(undef), @i4) # 'a', 1, 'b', 2, 'c', 3, undef, 4
zip(1..6, @i3 but fill(3), @i4 but fill('?'))
# 1,1,1,2,2,2,3,3,3,4,3,4,5,3,'?',6,3,'?'
Damian
