What's the fastest/most efficient way to count lines in Rebol?

Question

Given a string string, what is the fastest/most-efficient way to count lines therein? Will accept best answers for any flavour of Rebol. I\'ve been working unde

Answer 1

Here's the best simple non-parse version I can think of:

count-lines: function [text [string!]] [
    i: 1
    find-all text newline [++ i]
    i
]

It uses function and ++ from more recent versions of Rebol, and find-all from either R3 or R2/Forward. You could look at the source of find-all and inline what you find and optimize, but situations like this are exactly what we wrote find-all for, so why not use it?

Answer 2

Why no one came with the simplest solution I wonder :)

t: "abc^/de^/f^/ghi"
i: 0 until [i: i + 1 not t: find/tail t newline] i
== 4

Not sure about the performance but I think it's quite fast, as UNTIL and FIND are natives. WHILE could be used as well.

i: 1 while [t: find/tail t newline] [i: i + 1] i
== 4

Just need to check for empty string. And if it would be a function, argument series needs to be HEADed.

Answer 3

Do not know about performance, and the last line rule (r3).

>> length? parse "1^/2^/3" "^/"
== 3