Velocity Reviews > Problem understanding how closures work

Problem understanding how closures work

Tom Plunket
Guest
Posts: n/a

 12-12-2006
....at least, I think that I'm having a problem understanding the way
closures work.

I'm trying to define a function for an object which will take certain
objects from the parent scope at the time that function is defined.
For some reason, if I do this function definition in a loop, the
locals given by that function (is this a closure?) are changed each
iteration of the loop, whereas if the function definition is isn't
looped over, I get the behavior I desire. Can anyone provide any
insight for me?

thanks,
-tom!

First, the output:

Test 1 doesn't work the way I would expect:
Test 4 says, "Test 0"
Test 4 says, "Test 1"
Test 4 says, "Test 2"
Test 4 says, "Test 3"
Test 4 says, "Test 4"

....but test 2 does?
Test 0 says, "Test 0"
Test 1 says, "Test 1"
Test 2 says, "Test 2"
Test 3 says, "Test 3"
Test 4 says, "Test 4"

Next, the program:

class Test:
def __init__(self, name):
self.name = name

def DoCall(self):
self.ExternalCall(self.name)

# The first test.
def CreateTests1(count):
tests = []
for i in xrange(count):
name = 'Test %d' % i
t = Test(name)
tests.append(t)

def ExCall(text):
print '%s says, "%s"' % (name, text)

t.ExternalCall = ExCall

return tests

# The second test.
def CreateTests2(count):
tests = []
for i in xrange(count):
t = CreateTest(i)
tests.append(t)
return tests

def CreateTest(index):
name = 'Test %d' % index
t = Test(name)

def ExCall(text):
print '%s says, "%s"' % (name, text)

t.ExternalCall = ExCall
return t

print 'Test 1 doesn\'t work the way I would expect:'
for t in CreateTests1(5):
t.DoCall()

print '\n...but test 2 does?'
for t in CreateTests2(5):
t.DoCall()

Rob Williscroft
Guest
Posts: n/a

 12-12-2006
Tom Plunket wrote in news:(E-Mail Removed) in
comp.lang.python:

> ...at least, I think that I'm having a problem understanding the way
> closures work.
>
> I'm trying to define a function for an object which will take certain
> objects from the parent scope at the time that function is defined.
> For some reason, if I do this function definition in a loop, the
> locals given by that function (is this a closure?) are changed each
> iteration of the loop, whereas if the function definition is isn't
> looped over, I get the behavior I desire. Can anyone provide any
> insight for me?

> Test 1 doesn't work the way I would expect:
> Test 4 says, "Test 0"

> Test 4 says, "Test 4"
>

> def CreateTests1(count):
> tests = []
> for i in xrange(count):
> name = 'Test %d' % i
> t = Test(name)
> tests.append(t)
>
> def ExCall(text):
> print '%s says, "%s"' % (name, text)
>
> t.ExternalCall = ExCall
>
> return tests

"name" in the above code is bound to a an entry in "CreateTests1"'s
locals, and ExCall has a (hidden) reference to that locals, so
by the time ExCall is finally called the value associated
with "name" has been replaced by (count - 1).

The solution (as always) is to add another level of indirection:

def create_tests( count ):
def make( arg ):
def ExCall( text ):
print arg, text
return ExCall

tests = []

for i in range( count ):
name = i
t = Test( name )
t.ExternalCall = make( name )

In the above, every call to make() creates a new frame (a new set
of locals) and binds the value of the passed in "name" to the
name "arg" in this new frame, it will be this value that is
eventually printed.

There is a trick with default arguments that lets you do
what you want with a bit less faffing about:

>>> r = []
>>> for i in range(10):

def f( i = i ):
print i
r.append( f )

>>> for i in r:

i()

In this example the value of "i" is bound to the default argument
for the function "f" every time the def f() statments are executed.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

Gabriel Genellina
Guest
Posts: n/a

 12-12-2006
On 12 dic, 17:23, Tom Plunket <(E-Mail Removed)> wrote:

> ...at least, I think that I'm having a problem understanding the way
> closures work.
>
> I'm trying to define a function for an object which will take certain
> objects from the parent scope at the time that function is defined.

> def CreateTests1(count):
> tests = []
> for i in xrange(count):
> name = 'Test %d' % i
> t = Test(name)
> tests.append(t)
>
> def ExCall(text):
> print '%s says, "%s"' % (name, text)
>
> t.ExternalCall = ExCall
>
> return tests

name, inside ExCall, is a free variable. Python builds a closure
including the string whose name is "name" in the enclosing scope. Not
the *value* which happens to have at this momment. When you execute
ExCall, the reference to name yields its last, current, value.
If you want "the value at the moment the function is created" you can
use a default argument:

def ExCall(text, name=name): ...

Your second test works because you don't modify "name" between the
original definition and its execution.

--
Gabriel Genellina

Tom Plunket
Guest
Posts: n/a

 12-12-2006
Rob Williscroft wrote:

> "name" in the above code is bound to a an entry in "CreateTests1"'s
> locals, and ExCall has a (hidden) reference to that locals, so
> by the time ExCall is finally called the value associated
> with "name" has been replaced by (count - 1).

Ah, I got it. Thanks. Thanks too to Gabriel.

-tom!