From: Robert Klemme <bob.news@...>
Date: 2003-02-04T23:13:31+09:00
Subject: Re: Mismatch between documenation and implementation of Array#join()


"Gavin Sinclair" <gsinclair@soyabean.com.au> schrieb im Newsbeitrag
news:186173637337.20030205003517@soyabean.com.au...
> On Wednesday, February 5, 2003, 12:27:00 AM, Idan wrote:
>
> > Robert Klemme wrote:
>
> >>Hi,
> >>
> >>I'm wondering whether Array#join() has an error or the documentation is
> >>wrong. Documentation says:
> >>"Returns a string created by converting each element of the array to a
> >>string, separated by aSepString." This apparently does not include
> >>flatten(). But the behavior seems to be like flatten() was invoked
before:
> >>
> >>With Ruby version 1.6.7:
> >>
> >>irb(main):015:0* a=[[0,1],[22,33]]
> >>[[0, 1], [22, 33]]
> >>irb(main):016:0> a.join(";")
> >>"0;1;22;33"
> >>
> >>
> >>I would have expected a.join(";") to have returned "01;2233" instead.
> >>
> > I seem to think it is the correct behaviour, Array#join is just invoked
> > in each of the subarrays as well (Recursive behaviour, just as #to_s
> > calls #to_s on all of the array members), however, your wish might be
> > fulfilled this way: [...]
>
> That's a bad argument for the "correct behaviour" POV.  The doc
> doesn't specify recursive behaviour, so why do you expect it?  If you
> don't expect it, isn't the behaviour wrong, no matter how simply it
> can be described?

Or the documentation is in error.  However, I prefer to change the code to
actually match the documentation (see below).

> Array#to_s is actually specified (by 'ri') as returning 'arr.join',
> bringing us full circle.

Yeah, but the invocation used in to_s is equivalent to arr.join("") .
(Emphasis on empty string)

The problem I see is this: I can achieve the current behavior by issuing
a.flatten.join(";") which seems IMHO more appropriate than to use what I
regard as a workaround: a.map { |x| x.to_s }.join(';'). Using flatten makes
the recursive behavior explicit while the map solution will surely rise
question marks when people find this uncommented somewhere in code of other
people.

Regards

    robert