From: Stefano Crocco <stefano.crocco@...>
Date: 2009-02-05T21:10:47+09:00
Subject: Re: Array#choice always produce the same sequence

Alle Thursday 05 February 2009, Stefano Crocco ha scritto:
> I've just stumbled upon a strange behaviour of Array#choice (using ruby
> 1.8.7- p72). As far I understand, an_array.choice should be (almost) the
> same as an_array[rand(an_array.size)]. Thus, the two following pieces of
> code should be equivalent
>
> a = [1, 2, 3, 4]
> p 10.times.map{a.choice}.join
> p 10.times.map{a[rand(a.size)].join}
>
> However, this doesn't seem to be the case. In particular, something like:
>
> ruby -e 'a = [1, 2, 3, 4]; p 10.times.map{a.choice}.join'
>
> always give the same result, while

Sorry, hit the "Send" button too soon. Here's the full version

I've just stumbled upon a strange behaviour of Array#choice (using ruby
1.8.7- p72). As far I understand, an_array.choice should be (almost) the
same as an_array[rand(an_array.size)]. Thus, the two following pieces of
code should be equivalent

a = [1, 2, 3, 4]
p 10.times.map{a.choice}.join
p 10.times.map{a[rand(a.size)]}.join

However, this doesn't seem to be the case. In particular, something like:

ruby -e 'a = [1, 2, 3, 4]; p 10.times.map{a.choice}.join'

always give the same result, while

ruby -e 'a = [1, 2, 3, 4]; p 10.times.map{a[rand(a.size)]}.join'

gives a different result every time the command is executed (which, in my 
opinion, is the correct behaviour).

It seems that to get the correct behaviour from choice, you need to call srand 
before using it. Does anyone know why this is necessary with choice but not 
with rand? Is it the intended behaviour or a bug?

Thanks

Stefano