Mark Woodward
3/2/2008 6:42:00 AM
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