i think that returns a substring (ir a view onto the backing string). but
i am not sure. i did read a discussion somewhere saying that because of
this you should use bytestring(...) before passing a string to c. which is
all the evidence i have for my guess.
incidentally, match(...) has a method that takes the offset to start at as
an argument. so i can avoid s[i:end] and just pass i into match (i just
found this).
however, somewhat surprisingly, it also has the same problem.
andrew
On Tuesday, 21 July 2015 20:15:58 UTC-3, Yichao Yu wrote:
>
> On Tue, Jul 21, 2015 at 7:08 PM, Jameson Nash <vtj...@gmail.com
> <javascript:>> wrote:
> > does `copy` work? although `bytestring` also seems like a good method
> for
> > this also. it seems wrong to me also that `match` is making a copy of
> the
> > original string (if that is indeed what it is doing)
>
> Isn't it `s[i:end]` that is doing the copy?
>
> >
> > On Tue, Jul 21, 2015 at 6:57 PM andrew cooke <and...@acooke.org
> <javascript:>> wrote:
> >>
> >>
> >> string(bytestring(...)) seems to do it. would appreciate any more
> >> efficient solutions (and confirmation the analysis is correct - is this
> >> worth filing as an issue?)
> >>
> >>
> >> On Tuesday, 21 July 2015 19:33:05 UTC-3, andrew cooke wrote:
> >>>
> >>>
> >>> well, this was fun... the following code rapidly triggers the OOM
> killer
> >>> on my machine (julia 0.4 trunk):
> >>>
> >>> s = repeat("a", 1000000)
> >>> l = Any[]
> >>> r = r"^\w"
> >>>
> >>> for i in 1:length(s)
> >>> m = match(r, s[i:end])
> >>> push!(l, m.match)
> >>> end
> >>>
> >>> note that: (1) the regexp is only matching one character, so the array
> l
> >>> is at most a million characters long.
> >>>
> >>> what i think is happening (but this is only a guess) is that s[i:end]
> is
> >>> being passed though to the c level regexp library as a new string.
> the
> >>> result (m.match) is then a substring into that. because the substring
> is
> >>> kept around, the backing string cannot be collected. and so there's
> an n^2
> >>> memory use.
> >>>
> >>> ideally, i don't think a new copy of the string should be passed to
> the
> >>> regexp engine. maybe i am wrong?
> >>>
> >>> anyway, for now, if the above is right, i need some way to copy
> m.match.
> >>> as far as i can tell string() doesn't help. so what works? or am i
> wrong?
> >>>
> >>> thanks,
> >>> andrew
>
