Re: [bug-gawk] Why the length(seps) is one more than the return value of patsplit()?

[Manuel Collado]

El 08/03/2015 13:36, address@hidden escribió:
> Peng Yu <address@hidden> wrote:
> > The following code shows that the length(seps) is one more than the
> > return value of patsplit(). I thought that seps should have the same
> > as the number of elements in "a". Why it is not the case? Could
> > anybody help me understand this? Thanks.
> >
> > ~$ awk 'BEGIN { print patsplit("aAbBcC", a, /[a-z]/, seps); print
> > seps[1], seps[2], seps[3], length(seps); }'
> > 3
> > A B C 4
>
> Hi. Thanks for the report.  The fourth element is the null string. This
> may be a bug. I need to review the code and work with it a bit.  I'll
> reply again when I have more information.

IIRC, this was already discussed more than a year ago.

A record with N fields has N+1 potential separators, because there can be 
non null separators at both the begin and end of the record. At least for 
the default (space) separator.

Or perhaps it is the opposite: N separators may delimit N+1 fields.

Regards.
--
Manuel Collado - http://lml.ls.fi.upm.es/~mcollado