Comments on: Erlang FizzBuzz Showdown (pt2)

By: Guest

Guest — Mon, 14 Sep 2009 11:59:50 +0000

Interesting post. I have stumbled this for my friends. Hope others find it as interesting as I did.

By: JoelHughes

JoelHughes — Mon, 18 May 2009 14:07:23 +0000

Thanks, but as stated in Part 1: this was more to aid learning rather than using built in functionality (which I certainly will in future) – also the range function is modified to receive higher order functions, not sure that can be done with lists:seq?

By: JoelHughes

JoelHughes — Mon, 18 May 2009 14:04:21 +0000

The “From < To” is simply to stop an infinite loop if the function was called backwards e.g. range(5,1).

By: Guest

Guest — Sat, 16 May 2009 16:31:52 +0000

lists:seq(1,100).

By: iWantToKeepAnon

iWantToKeepAnon — Sat, 16 May 2009 03:55:29 +0000

Ah, ok there is the list comprehension. Very good. But the guard “From < To” still isn't (AFAIK) necessary. Erlang matches your function clauses top to bottom and the “range(To,To,F)” clause would break you out of your recursion.

What is the guard for?

By: joelhughes

joelhughes — Fri, 15 May 2009 14:23:29 +0000

Thanks Matt, I didn't know that short cut – pretty handy!

By: Matt Brubeck

Matt Brubeck — Thu, 14 May 2009 23:44:20 +0000

Just a minor comment - this function:


  combine(To) ->
    F = (fun(I) ->  fizzbuzz(I) end),
    lists:map(F,range(1,To,[])).

could be written a little more clearly and concisely as:


  combine(To) ->
    lists:map(fun fizzbuzz/1, range(1,To,[])).

By: Joel Hughes’s Blog » Erlang FizzBuzz Showdown (Pt 1)

Joel Hughes’s Blog » Erlang FizzBuzz Showdown (Pt 1) — Thu, 14 May 2009 11:43:12 +0000

[...] Part 2 >> [...]