在Rebol中计算行数最快/最有效的方法是什么? [英] What's the fastest/most efficient way to count lines in Rebol?

问题描述

给出一个字符串string，计算其中行数的最快/最有效方法是什么?对于任何口味的Rebol都将接受最佳答案.我一直在假设parse [some [thru]]组合是遍历字符串的最快方法，但后来我不确定，因此转向SO:

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
]

或者:

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

柜台又如何呢?重复的效率如何?

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
        ]
    ]
]

更新:行数遵循文本编辑器"原则:

Update: line count goes by the Text Editor principle:

一个空文档的行数仍为1.所以:

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

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

推荐答案

增强的PARSE版本，由BrianH建议:

Enhanced PARSE version, as suggested by BrianH:

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

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