 What am I missing with this closure example?

Hello!

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

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

x = 0
two = f “two”
2
I thought that x is shared between one and two but they don’t seem to.
Or am I missing something?

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

The “`Stok’ disclaimers” apply.

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

“Sam K.” [email protected] 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.

[email protected] wrote:

one, two = f
2
2

Aha~
Now I see the difference.
Thanks everyone.

Sam

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 K.” [email protected] wrote in message
news:[email protected]

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