When Closure get external variable's value?

[Huayang Xia]

What will the following piece of code print? (10 or 15)

def testClosure(max Index) :

def closureTest():
return maxIndex

maxIndex += 5

return closureTest()

print testClosure(10)

My question is when the closure function gets value for maxindex? Run
time or compile time?

Thanks.

[Huayang Xia]

It will print 15. The closure gets the value at run time.

Could we treat closure as part of the external function and it shares
the local variable with its holder function?

[Jussi Salmela]

Huayang Xia kirjoitti:

It will print 15. The closure gets the value at run time.

Could we treat closure as part of the external function and it shares
the local variable with its holder function?

I don't quite get what you are trying to tell us but if you think that
in your example code:

def testClosure(max Index) :

def closureTest():
return maxIndex

maxIndex += 5

return closureTest()

print testClosure(10)

you are returning a callable function you are all wrong. This can be
easily seen by:

>>type(testClos ure(10))

15
<type 'int'>

The mistake is that you shouldn't return closureTest() but closureTest
instead. The correct way would be:

>>def testClosure2(ma xIndex):

def closureTest():
return maxIndex
maxIndex += 5
return closureTest

>>f2 = testClosure2(10 )

<function closureTest at 0x00D82530>

>>type(f2)

<type 'function'>

>>print f2()

15

Cheers,
Jussi

[Huayang Xia]

Thanks for the clarification.

But my question is:

When does the closure get the value of the maxIndex in the following
code snippet?

def testClosure(max Index) :

def closureTest():
return maxIndex

maxIndex += 5

return closureTest()

print testClosure(10)
I thought it should be 10 instead of 15. That was wrong.

After several tests, I found maxIndex is, though, local to
testClosure() but is external to the closureTest(). closureTest() gets
the value of maxIndex at run time. So that it's 15 instead of 10. The
following snippet will verify that further:

def testClosure1(ls t):

def closureTest():
lst.append(lst[-1]+1)

lst.append(lst[-1]+1)
return closureTest()

alist = [1]
testClosure1(al ist)
alist.append(3)
testClosure1(al ist)

The 'lst' in function testClosure1() and the closure closureTest() are
same thing as alist. So everything is dynamic. Variable's value is
determined at run time.

[Fredrik Lundh]

Huayang Xia wrote:

When does the closure get the value of the maxIndex in the following
code snippet?

def testClosure(max Index) :

def closureTest():
return maxIndex

maxIndex += 5

return closureTest()

print testClosure(10)
I thought it should be 10 instead of 15. That was wrong.

free variables in an inner scope bind to variables in the outer scope,
not objects.

if you want to bind to objects, use explicit binding:

def closureTest(max Index=maxIndex) :
return maxIndex

</F>

[Huayang Xia]

That is a really concise and precise answer. Thanks.

So the object binding can only happen explicitly at the closure
declaration argument list(non-free variable).

On Dec 19, 10:37 am, Fredrik Lundh <fred...@python ware.comwrote:

Huayang Xia wrote:

When does the closure get the value of the maxIndex in the following
code snippet?

def testClosure(max Index) :

def closureTest():
return maxIndex

maxIndex += 5

return closureTest()

print testClosure(10)

I thought it should be 10 instead of 15. That was wrong.free variables in an inner scope bind to variables in the outer scope,

not objects.

if you want to bind to objects, use explicit binding:

def closureTest(max Index=maxIndex) :
return maxIndex

</F>

[Marc 'BlackJack' Rintsch]

In <11************ **********@t46g 2000cwa.googleg roups.com>, Huayang Xia
wrote:

That is a really concise and precise answer. Thanks.

So the object binding can only happen explicitly at the closure
declaration argument list(non-free variable).

That's no declaration that's a definition and it happens at runtime! It's
executed every time the outer function is called and executed and creates
a function object.

As far as I can see you don't create a closure BTW. You are *calling*
that inner function and return the *result* of that call.

Ciao,
Marc 'BlackJack' Rintsch

[Huayang Xia]

I'm confused. What is the definition of closure.

I'm not sure if it's correct, I get the definition from wikipedia:

"A closure typically comes about when one function is declared entirely
within the body of another, and the inner function refers to local
variables of the outer function. At runtime, when the outer function
executes, a closure is formed. It consists of the inner function's code
and references to any variables in the outer function's scope that the
closure needs."

I agree it is not declaration, it's definition. However it's closure
based on the above definition. It uses free variable. Or you mean it's
a closure only when the outer function returns it and be exposed to
external world?

The code snippet was just for test purpose. My question was how the
free variable inside inner function (the closure) binds with object.
That was answered by Fredrik perfectly.

[Bruno Desthuilliers]

Huayang Xia a écrit :

I'm confused. What is the definition of closure.

I'm not sure if it's correct, I get the definition from wikipedia:

"A closure typically comes about when one function is declared entirely
within the body of another, and the inner function refers to local
variables of the outer function. At runtime, when the outer function
executes, a closure is formed. It consists of the inner function's code
and references to any variables in the outer function's scope that the
closure needs."

You skipped the first and most important sentence:
"In programming languages, a closure is a function that refers to free
variables in its lexical context."

IOW, a closure is a function that carry it's own environment. In the
following code, the function returned by make_adder is a closure :

def make_adder(addi ng):
def adder(num):
return num + adding
return adder
add_three = make_adder(3)
print add_three(4)
=7

I agree it is not declaration, it's definition. However it's closure
based on the above definition. It uses free variable.

Actually, it uses a variable defined in the enclosing scope. But as long
as it's also executed in the same enclosing scope, it's just a nested
function.

Or you mean it's
a closure only when the outer function returns it and be exposed to
external world?

Bingo.

[Fredrik Lundh]

Bruno Desthuilliers wrote:

>You skipped the first and most important sentence:

"In programming languages, a closure is a function that refers to free
variables in its lexical context."

IOW, a closure is a function that carry it's own environment.

in contrast to functions that don't know what environment they belong
to, you mean ? can you identify such a construct in Python ?

</F>