How to make inner functions see variables of enclosing function?

Question

Possible Duplicate:
Read/Write Python Closures

In the following function, the internal function does not modify the argument but just modifies the copy.

def func():
  i = 3
  def inc(i):
    i = i + 3
  print i
  inc(i)
  inc(i)
  print i

func()

Is it possible to avoid repeated code and put that inside a function in python? I tried the following too but it throws error UnboundLocalError: local variable 'i' referenced before assignment

def func():
  i = 3
  def inc():
    i = i + 3
  print i
  inc()
  inc()
  print i

func()

It is one of my favorite interview questions. I suggest you read about variable visibility in Python. — lig, Dec 20 '11 at 15:41
possible duplicate of [Read/Write Python Closures](http://stackoverflow.com/q/2009402/395760) — , Dec 20 '11 at 15:42

score 3 · Accepted Answer · answered Dec 20 '11 at 15:49

In python 3 you'd do this:

def func():
  i = 3
  def inc():
    nonlocal i
    i = i+3
  print(i)
  inc()
  inc()
  print(i)

func()

In python 2.x using global doesn't work because the variable is in an outer scope, but it's not global. Hence, you'll need to pass the variable as an argument.

This is the problem that PEP 3104 addresses.

score 2 · Answer 2 · answered Dec 20 '11 at 15:38

2

what about:

def func():
    i = 3
    def inc(i):
        return i + 3
    print i
    i = inc(i)
    i = inc(i)
    print i

func()

answered Dec 20 '11 at 15:38

hlt

6,219
3
23
43

score 0 · Answer 3 · answered Dec 20 '11 at 15:51

0

In Python 3 you can use nonlocal.

>>> def func():
    i = 3
    def inc():
        nonlocal i
        i += 3
    print(i)
    inc()
    inc()
    print(i)

>>> func()
3
9
>>>

answered Dec 20 '11 at 15:51

yak

8,851
2
29
23

How to make inner functions see variables of enclosing function?

3 Answers3