On 5/12/05, Frank Ellermann <nobody(_at_)xyzzy(_dot_)claranet(_dot_)de> wrote:
Bruce Lilly wrote:
e.g. the unused
construct. After some thought, you may realize that that's
equivalent to *5foo
I'd expect 0, 3, 4, or 5 foo. If what you say is true it has
"a high astonishing factor", one of the taboos in my religion.
Same idea for 3*5*1(foo) => 3*5(*1(foo)) => 0, 3, 4, or 5 foo.
That's what I expected, how do you get *5(foo) ? I don't see
1*2(foo). But I love syntax puzzles ;-)
I get *5foo too - it's one of the "canonicalizations" that my parser does.
a = 3*5[foo]
; foo UNDEFINED
a = *5foo
I think of it this way: you're saying you have between 3 and 5 [foo]s.
Each [foo] may be empty or may be a foo. So, you can get any number
between 0 and 3 inclusive by having some empty [foo]s and some
If I wanted 0, 3, 4 or 5 foos I think I'd say [3*5foo].