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in reply to Re^2: Lexical closures
in thread Lexical closures

The real question is, what do your Scheme and Common Lisp code samples look like?

The Perl behaviour is precisely what I would expect. It generates three closures, each capturing a value of $i. So your first foreach block reduces to: 

my @flist = ( sub {0 * $_[0]}, sub {1 * $_[0]}, sub {2 * $_[0]});

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It's obvious, then, that 0 2 4 is the correct output of 
foreach my $f (@flist)
{
    print $f->(2), "\n";
}

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Re^4: Lexical closures
by mpeever (Friar) on Oct 25, 2008 at 20:07 UTC

    OK, I took some time to mock out some Scheme: here are both behaviours you've seen: 

    (define cclosures 
  (lambda (values)
    (cond ((null? values) '())
          (else (cons (lambda (x) (* (car values) x))
                      (cclosures (cdr values)))))))

(define cclosures2
  (let ((val -1))
    (lambda (values)
      (cond ((null? values) '())
            (else
             (begin
               (set! val (car values))
               (cons (lambda (x) (* val x))
                     (cclosures2 (cdr values)))))))))

(define clprint
  (lambda (closures)
    (map (lambda (fn) (fn 2)) closures))) 

> (clprint (cclosures '(0 1 2)))
(0 2 4)
> (clprint (cclosures2 '(0 1 2)))
(4 4 4)

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    cclosures uses the equivalent ofdeclaring my $i inside the foreach loop: it defines a new variable called val for every iteration.

    cclosures2 uses the equivalent of declaring my $i outside the foreach loop: val gets reassigned with every iteration.

    Notice the closure closes over the variable, not the value. So when the variable is reassigned, the value inside the closure changes too.

    HTH

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