Hi,
I've tried searching for this on the forum and couldn't find an answer.
Apologies if it's already out there.
I've converted the digits of an integer into elements of an array, like
so:
x = some_integer
y = "#{x}".scan(/./)
Better
y = x.to_s.scan /./
You can also do this numerically
irb(main):008:0> x = 113453
=> 113453
irb(main):009:0> y =
=>
irb(main):010:0> while x > 0; x, b = x.divmod 10; y.unshift b; end
=> nil
irb(main):011:0> y
=> [1, 1, 3, 4, 5, 3]
respectively
irb(main):015:0> x = 113453
=> 113453
irb(main):016:0> y =
=>
irb(main):017:0> while x > 0; x, b = x.divmod 10; y.unshift b.to_s; end
=> nil
irb(main):018:0> y
=> ["1", "1", "3", "4", "5", "3"]
or even
irb(main):030:0> x = 113453
=> 113453
irb(main):031:0> y =
=>
irb(main):032:0> while x > 0; y.unshift((x, b = x.divmod 10).last.to_s); end
=> nil
irb(main):033:0> y
=> ["1", "1", "3", "4", "5", "3"]
Now, after using y for my nefarious purposes, I would like to re-convert
the elements (still containing one digit each) into an integer.
You make me curious...
I don't mind writing a routine for it, but was wondering if there's some
magical way to do it quicker 
No magic needed
z = y.join.to_i
Kind regards
robert
路路路
On Thu, Oct 11, 2012 at 8:14 PM, Anna Baas <lists@ruby-forum.com> wrote:
--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/
