python - How do lexical closures work? -

while investigating problem had lexical closures in javascript code, came along problem in python:

flist = []  in xrange(3):     def func(x): return x *     flist.append(func)  f in flist:     print f(2) 

note example mindfully avoids lambda. prints "4 4 4", surprising. i'd expect "0 2 4".

this equivalent perl code right:

my @flist = ();  foreach $i (0 .. 2) {     push(@flist, sub {$i * $_[0]}); }  foreach $f (@flist) {     print $f->(2), "\n"; } 

"0 2 4" printed.

can please explain difference ?

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update:

the problem is not i being global. displays same behavior:

flist = []  def outer():     in xrange(3):         def inner(x): return x *         flist.append(inner)  outer() #~ print   # commented because causes error  f in flist:     print f(2) 

as commented line shows, i unknown @ point. still, prints "4 4 4".

python behaving defined. three separate functions created, each have closure of environment they're defined in - in case, global environment (or outer function's environment if loop placed inside function). problem, though - in environment, i mutated, , closures refer same i.

here best solution can come - create function creater , invoke that instead. force different environments each of functions created, different i in each one.

flist = []  in xrange(3):     def funcc(j):         def func(x): return x * j         return func     flist.append(funcc(i))  f in flist:     print f(2) 

this happens when mix side effects , functional programming.