What am I missing with this closure example?

ruby-talk
1

Hello!

Since my last question about
closure(http://groups.google.com/group/comp.lang.ruby/browse_frm/thread/0d7e258107e1683e/12e182a6ff493232?hl=en#12e182a6ff493232)
I am testing my understanding making some examples.

Here's some code:

def f name
  x = 0

  case name
    when "one"
      lambda {x += 1; puts x}
    when "two"
      lambda {puts x}
  end
end

one = f "one"
two = f "two"

one.call
two.call
one.call
two.call

Result:

1
0
2
0

I expected the following.
1
1
2
2

I thought that x is shared between one and two but they don't seem to.
Or am I missing something?

Thanks.

Sam

2

Like konsu states, you have two different closures. Here's a little
rewrite of your code that works with one closure.

def f
  x = 0
  one = lambda {x += 1; puts x}
  two = lambda {puts x}
  [one, two]
end

one, two = f

one.call
two.call
one.call
two.call

=> with results of
1
1
2
2

- Hitesh
http://www.jasani.org/

3

"Sam Kong" <sam.s.kong@gmail.com> writes:

def f name
x = 0

case name
  when "one"
    lambda {x += 1; puts x}
  when "two"
    lambda {puts x}
end
end

one = f "one"
two = f "two"

The assignment "x = 0" gets executed every time you call f, creating a new x
each time and initializing that new x to zero.

As far as I can tell, Ruby has no static scope, only dynamic scope, which,
if I'm right, means there's no straightforward way to do what you're trying.

In Perl, for instance, you could do this:

{
    my $x = 0;
    sub f
    {
        my $name = shift;
        return sub { $x += 1; print "$x\n"; } if ($name eq 'one');
        return sub { print "$x\n"; } if ($name eq 'two');
    }
}
my $one = f('one');
my $two = f('two');
$one->();
$two->();
$one->();
$two->();

And achieve the result you expected, but in Ruby, I don't know how to achieve
something similar. As a practical matter, I would just create an object with
an attribute and use methods on that object in lieu of the closures.

4

hello,

f("two") creates a lambda that sees 'x' as it was at the time of its
creation. zero that is.

the 'x' that f("one") uses is in a different closure.

konstantin

"Sam Kong" <sam.s.kong@gmail.com> wrote in message
news:1141155782.675003.313420@e56g2000cwe.googlegroups.com...

···

Hello!

Since my last question about
closure(http://groups.google.com/group/comp.lang.ruby/browse_frm/thread/0d7e258107e1683e/12e182a6ff493232?hl=en#12e182a6ff493232\)
I am testing my understanding making some examples.

Here's some code:

def f name
x = 0

case name
when "one"
lambda {x += 1; puts x}
when "two"
lambda {puts x}
end
end

one = f "one"
two = f "two"

one.call
two.call
one.call
two.call

Result:

1
0
2
0

I expected the following.
1
1
2
2

I thought that x is shared between one and two but they don't seem to.
Or am I missing something?

Thanks.

Sam

5

Each time you call f you get a different x, so the lambda's x are bound to different variables. Does this do what you expect:

def f
  x = 0

   [lambda {x += 1; puts x}, lambda {puts x}]
end

one, two = f

one.call
two.call
one.call
two.call

Mike

···

On 28-Feb-06, at 8:58 PM, Sam Kong wrote:

Hello!

Since my last question about
closure(http://groups.google.com/group/comp.lang.ruby/browse_frm/thread/0d7e258107e1683e/12e182a6ff493232?hl=en#12e182a6ff493232\)
I am testing my understanding making some examples.

Here's some code:

def f name
  x = 0

  case name
    when "one"
      lambda {x += 1; puts x}
    when "two"
      lambda {puts x}
  end
end

one = f "one"
two = f "two"

one.call
two.call
one.call
two.call

Result:

1
0
2
0

I expected the following.
1
2

I thought that x is shared between one and two but they don't seem to.
Or am I missing something?

--

Mike Stok <mike@stok.ca>
http://www.stok.ca/~mike/

The "`Stok' disclaimers" apply.

6

Aha~
Now I see the difference.
Thanks everyone.

Sam

···

hitesh.jasani@gmail.com wrote:

Like konsu states, you have two different closures. Here's a little
rewrite of your code that works with one closure.

def f
  x = 0
  one = lambda {x += 1; puts x}
  two = lambda {puts x}
  [one, two]
end

one, two = f

one.call
two.call
one.call
two.call

=> with results of
1
1
2
2