How to allow exec to access free variables in nested method in python?

Question

I have to access free variables in exec in nested method in python(2.6)

def outerF():
    la = 5
    def innerF():
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

The expected result is 5 but I am getting the following error

NameError: name 'la' is not defined

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal. — Martijn Pieters, Jan 02 '19 at 10:28
Moreover, `la` is **not a free variable**. Variables are only marked as free when the compiler sees that there is a nested function that uses the name, and there is no such use here at compile time. By the time `exec` is executed it's too late to change this. — Martijn Pieters, Jan 02 '19 at 10:33
Closure : It is a function object that remembers values in enclosing scopes even if they are not present in memory. To solve this you need to use **nonlocal** keyword You can check use of **nonlocal** and details about **Closure** in following link for more info https://www.learnpython.org/en/Closures — AviatorX, Jan 02 '19 at 10:39

sahasrara62 · Answer 1 · 2019-01-02T10:49:55.847

0

for python 2.6

def outerF():
    la={'y': 5} # exec take argu as dict and in 2.6 nonlocal is not working, 
    # have to create dict element and then process it 
    def innerF():

        str1 = "print({})".format(la['y'])
        exec (str1) in globals(),locals()
    innerF()
outerF()

edited Jan 02 '19 at 10:49

answered Jan 02 '19 at 10:21

sahasrara62

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Care to add some explanation to your undocumented answer? I didn't downvote though. – Sheldore Jan 02 '19 at 10:22
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You are now accessing a global variable, not the free variable in the closure of inner – juanpa.arrivillaga Jan 02 '19 at 10:23
How to do it without making la global? – Vicky Gupta Jan 02 '19 at 10:26
@Bazingaa hope this is explanatory enough – sahasrara62 Jan 02 '19 at 10:40
@juanpa.arrivillaga updated the code using the global or nonlocal way – sahasrara62 Jan 02 '19 at 10:42
@VickyGupta usng nonlocal – sahasrara62 Jan 02 '19 at 10:44
It is python 2.6 @prashantrana – Vicky Gupta Jan 02 '19 at 10:45
@VickyGupta updated for 2.6 – sahasrara62 Jan 02 '19 at 10:50

score 0 · Answer 2 · edited Jan 02 '19 at 10:32

0

Use nonlocal to access outer function's variable in nested functions as: Python Global, Local and Nonlocal variables

def outerF():
    la = 5
    def innerF():
        nonlocal la    #it let the la variable to be accessed in inner functions.
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

edited Jan 02 '19 at 10:32

khelwood

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answered Jan 02 '19 at 10:22

ThunderMind

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It is python 2.6 – Vicky Gupta Jan 02 '19 at 10:37
if you don't want to use Global, you might find your answer here than :) `https://stackoverflow.com/questions/8447947/is-it-possible-to-modify-variable-in-python-that-is-in-outer-but-not-global-sc/8448029` – ThunderMind Jan 02 '19 at 10:48

score 0 · Answer 3 · edited Jan 02 '19 at 12:39

0

Please try using nonlocal statement for your variable. Like this:

def outerF():
    la = 5
    def innerF():
        nonlocal la
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

More info could be found here: Python Global, Local and Nonlocal variables

edited Jan 02 '19 at 12:39

ThunderMind

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answered Jan 02 '19 at 10:24

Alexander Lyapin

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It is giving error: nonlocal la ^ SyntaxError: invalid syntax – Vicky Gupta Jan 02 '19 at 10:25
@VickyGupta nonlocal isn't available in Python 2. If you're using an old version of Python, please specify that in your question. – khelwood Jan 02 '19 at 10:27
@VickyGupta, try to avoid global vars as much as possible. May be passsing it as an argument with the default value or using *args is better idea. – Alexander Lyapin Jan 02 '19 at 10:32

How to allow exec to access free variables in nested method in python?

3 Answers3