Re: srfi-1 take and drop seriously broken

[dsmich]

---- Panicz Maciej Godek <address@hidden> wrote: 
> 2016-11-20 11:42 GMT+01:00 Jan Synáček <address@hidden>:
> 
> >
> > >> Please, tell me that this is just a mistake... This can't be true. I
> > >> still can't believe it. This is from 2.0.11. Please, tell me that the
> > >> implementation is fixed in 2.2.
> > >>
> > >> Yours truly puzzled,
> > >
> > >
> > > I don't know why you find it so puzzling. You can't take or drop
> > something
> > > that "isn't there" (you can't take a car or cdr from an empty list as
> > well,
> > > although e.g. in the language of "The Little Prover" (car '()) and (cdr
> > '())
> > > both evaluate to '() to assure their totality). If you need, you can
> > define
> > > your own variants that take/drop at most n elements of list.
> >
> > Not only that you "can", it's also IMHO a fool-proof implementation
> > and I can't see any reason why it should behave differently.
> >
> 
> Because someone might think that if he took 7 elements, then he has 7
> elements, so it is good that he knows earlier that this is not the case. I
> don't see a point in referring to Haskell documentation when discussing
> Scheme functions, though (if you try Racket, you'll note that although its
> take has a reversed order of arguments compared to srfi-1, it still doesn't
> allow to take or drop a positive number of elements from an empty list)
> 
> I agree though that the srfi-1 document isn't explicit enough about this
> point.

Yes.   Srfi-1 ( http://srfi.schemers.org/srfi-1/srfi-1.html ) says that drop 
"is exactly equivalent to performing i cdr operations on x", and notes that for
car and cdr "it is an error to apply them to the empty list"

Also for take and drop: "For a legal i, take and drop partition the list in a 
manner which can be inverted with append"
Implying that not all i are "legal".

-Dale