From: Mark Woodward <markonlinux@...>
Date: 2008-03-02T15:44:55+09:00
Subject: Re: What should this print?

Hi Bouquet,

On Sun, 02 Mar 2008 05:04:47 GMT
Bouquet <news@soma.bpa.nu> wrote:

> Hi,
> 
> I was expecting the code below to print 1, 1,
> but I get 1, nil.
> 
> old = nil
> [1,1].each do |n|
>    if n != old
>      x = 1
>      old = n
>    end
>    p x
> end
> 
> Thanks.

1.	old = nil
2.	[1,1].each do |n|
3.	   if n != old
4.	     x = 1
5.	     old = n
6.	   end
7.	   p x
8.	end


old was declared *outside* the each block at (1), so its remembered in,
as well as after the each block.

old = nil
[1,1].each do |n|
   if n != old
     x = 1
     old = n
   end
   p x
end
puts "old outside the block: #{old}"


x declared *inside* the each block at (4), *only* when the if condition
is true! The first iteration (the first 1 in [1,1] at (2)), n doesn't
equal old (n=1, old=nil) so the if condition is true. So x is initiated
to 1, old becomes 1.
   p x   #=> prints 1
Now, (and this is a newbies understanding), at the end of this first
iteration at (8) x is forgotten. ie not available during the second
iteration.

If we start the second iteration, n=1 (the second 1 in [1,1] at (2)
and old (which *is* still available because it was declared outside the
each block) = 1. So if n=1 and old=1, we would never enter the if
statement at (3) and x would not be initiated. Hence p as nil.


Try this:

    old = nil
    [1,1].each do |n|
       if n != old
         x = 1
         old = n
       else
         x = "what am I?"
       end
       p x
    end


and this:


    old = nil
    x = nil
    [1,1].each do |n|
       if n != old
         x = 1
         old = n
       end
       p x
    end


cheers,


-- 
Mark