On Thursday 24 March 2016, Bernhard Voelker wrote:
> On 03/24/2016 04:28 PM, Маренков Евгений wrote:
> > I have recently noticed an apparent bug in 'seq'. If one runs
> > seq -w 2 1 10
> > everything works fine.
> > But
> > seq -w 2 0 10
> > falls into endless loop ...
>
> Thanks for the report.
> However, I don't think this is a bug but more a misunderstanding
> on your side how seq works. It's clearly documented that the
> second number (if 3 are given) is the increment. As you passed
> 0 as value, the target 10 is never reached, and therefore seq
> continues to produce output. Thus, it exactly does what you
> told it to do. What else would you expect?
This sounds all true, however then these one should also run forever:
$ seq 10 0 2
Man page says:
INCREMENT is usually positive if FIRST is smaller than LAST,
and INCREMENT is usually negative if FIRST is greater than LAST.
This implicates IMO that seq should try to count _down_ if FIRST > LAST
and INCREMENT=0
Moreover I'd say this one does not need to loop endless:
$ seq 0 0 0
BTW maybe it would be an improvement to handle epsilon related endless
loops like this
$ seq 1 0.00000000000000000000000000000001 2
cu,
Rudi
