[racket] substring without the copy

Unfortunately that eventually does require running the true substring operation. All of my substrings are short-lived values that are used immediately, so being lazy doesn't help much. My issue is that Im running substring several million times to generate very short-lived values that are almost immediately ready to be collected.


substring/shared does seem to be what I want, although for some reason using that instead of substring is actually much  (~30%) slower. My initial guess as to why is TR contract boundary shenanigans, although that also seems odd given that the only contract that should be generated by requiring substring/shared is string?.

On Fri, Apr 4, 2014 at 4:12 PM, John Clements <clements at brinckerhoff.org>
wrote:

> On Apr 3, 2014, at 7:08 PM, Spencer Florence <spencer at florence.io> wrote:
>> Is there any way to perform `substring` without creating a new string? I'm working with lots of very large and (for all intents and purposes) immutable strings and would like to avoid the extra allocations and copy time.
> Seems like it would be fairly easy to roll this yourself:
> #lang racket
> (require rackunit)
> ;; represent a lazy substring operation:
> (struct lazysubstring (str start end))
> ;; force a lazy substring
> (define (lss-force s)
>   (substring (lazysubstring-str s)
>              (lazysubstring-start s)
>              (lazysubstring-end s)))
> ;; get the length of a lazy substring:
> (define (lss-len s)
>   (- (lazysubstring-end s) (lazysubstring-start s)))
> ;; create one:
> (define source-text "othho tot stnh ontuh .nt")
> (define lss1 (lazysubstring source-text 3 9))
> (check-equal? (lss-len lss1) 6)
> Obviously, you have to “roll your own” for any operation that you want to perform that retains the laziness.
> John
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