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 under the assumption that the parse [some [thru]] combination was the fastest way to traverse a string, but then I don't know that for certain, hence turning to SO:

count-lines: func [string [string!] /local count][
    parse/all string [
        (count: 1) some [thru newline (count: count + 1)]
    ]
    count
]

Or:

count-lines: func [string [string!] /local count][
    count: 0
    until [
        count: count + 1
        not string: find/tail string newline
    ]
    count
]

And how about counters? How efficient is repeat?

count-lines: func [string [string!]][
    repeat count length? string [
        unless string: find/tail string newline [
            break/return count
        ]
    ]
]

Update: line count goes by the Text Editor principle:

An empty document still has a line count of one. So:

>> count-lines ""
== 1
>> count-lines "^/"
== 2

Answer 1

count-lines: func [
    str
    /local sort-str ][
sort-str: sort join str "^/"
1 + subtract index? find/last sort-str "^/" index? find sort-str "^/"
]

Answer 2

Enhanced PARSE version, as suggested by BrianH:

i: 1 ; add one as TextMate
parse text [any [thru newline (++ i)]]
print i

Answer 3

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?