Question about nested scopes

[Miki Tebeka]

Hello,

Can anyone explain why:

def make_inc(n): s = n
def inc(i):
s += i
return s
return inc
i = make_inc(3)
i(2)
Traceback (most recent call last):
File "<pyshell#3 6>", line 1, in -toplevel-
i(2)
File "<pyshell#3 4>", line 4, in inc
s += i
UnboundLocalErr or: local variable 's' referenced before assignment
But: def make_inc(n): s = [n]
def inc(i):
s[0] += i
return s[0]
return inc
i = make_inc(2)
i(2) 4 i(2)

6

Thanks.
Miki

[Emile van Sebille]

"Miki Tebeka" <te****@cs.bgu. ac.il> wrote in message
news:33******** *************** ***@posting.goo gle.com...

Hello,

Can anyone explain why:

def make_inc(n): s = n
def inc(i):
s += i
return s
return inc [error snipped]
But: def make_inc(n): s = [n]
def inc(i):
s[0] += i
return s[0]
return inc

For the same reason as

s=1
s is 1 True s+=1
s is 2

True

Assignment to from within a function creates a local variable. Modification
of a mutable does not.

--
Emile van Sebille
em***@fenx.com

[Holger Krekel]

Miki Tebeka wrote:

Hello,

Can anyone explain why:

def make_inc(n): s = n
def inc(i):
s += i
return s
return inc
i = make_inc(3)
i(2)
Traceback (most recent call last):
File "<pyshell#3 6>", line 1, in -toplevel-
i(2)
File "<pyshell#3 4>", line 4, in inc
s += i
UnboundLocalErr or: local variable 's' referenced before assignment

you cannot change name-bindings in outer scopes.
the compiler probably thinks because of 's+=i' that s is a
local variable to 'inc', the inner function. but upon execution
it notices that it is not initialized and raises the usual exception.

However,

But: def make_inc(n):

s = [n]
def inc(i):
s[0] += i
return s[0]
return inc

here you don't modify the name-binding (s references a list
all the time) but you mutate the list.

holger

[Miki Tebeka]

Hello Emile,

Assignment to from within a function creates a local variable. Modification
of a mutable does not.

Ok, thanks.

Miki