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

This question already has an answer here:

Modify bound variables of a closure in Python

7 answers

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

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

marked as duplicate by Martijn Pieters♦ python
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Jan 2 at 10:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal.

– Martijn Pieters♦
Jan 2 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 2 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 learnpython.org/en/Closures

– AviatorX
Jan 2 at 10:39

add a comment |

This question already has an answer here:

Modify bound variables of a closure in Python

7 answers

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

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

marked as duplicate by Martijn Pieters♦ python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 2 at 10:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal.

– Martijn Pieters♦
Jan 2 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 2 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 learnpython.org/en/Closures

– AviatorX
Jan 2 at 10:39

add a comment |

This question already has an answer here:

Modify bound variables of a closure in Python

7 answers

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

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

This question already has an answer here:

Modify bound variables of a closure in Python

7 answers

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

This question already has an answer here:

Modify bound variables of a closure in Python

7 answers

python python-2.x

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

edited Jan 2 at 10:43

asked Jan 2 at 10:16

Vicky Gupta

10118

asked Jan 2 at 10:16

Vicky Gupta

10118

asked Jan 2 at 10:16

Vicky Gupta

10118

marked as duplicate by Martijn Pieters♦ python
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Jan 2 at 10:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martijn Pieters♦ python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 2 at 10:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal.

– Martijn Pieters♦
Jan 2 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 2 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 learnpython.org/en/Closures

– AviatorX
Jan 2 at 10:39

add a comment |

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal.

– Martijn Pieters♦
Jan 2 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 2 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 learnpython.org/en/Closures

– AviatorX
Jan 2 at 10:39

You can't. Not in Python 2, where there is no concept of marking variables you bind to as nonlocal.

– Martijn Pieters♦
Jan 2 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 2 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 learnpython.org/en/Closures

– AviatorX
Jan 2 at 10:39

add a comment |

3 Answers
3

active

oldest

votes

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 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

add a comment |

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 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

1

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

1

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

|
show 3 more comments

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 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

add a comment |

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 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

add a comment |

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 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

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 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

edited Jan 2 at 10:32

khelwood

31.9k74465

edited Jan 2 at 10:32

khelwood

31.9k74465

edited Jan 2 at 10:32

khelwood

31.9k74465

answered Jan 2 at 10:22

ThunderMind

639214

answered Jan 2 at 10:22

ThunderMind

639214

answered Jan 2 at 10:22

ThunderMind

639214

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

add a comment |

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

It is python 2.6

– Vicky Gupta
Jan 2 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 2 at 10:48

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 2 at 10:48

add a comment |

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 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

1

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

1

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

|
show 3 more comments

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 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

1

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

1

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

|
show 3 more comments

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 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

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 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

edited Jan 2 at 10:49

answered Jan 2 at 10:21

prashant rana

945920

answered Jan 2 at 10:21

prashant rana

945920

answered Jan 2 at 10:21

prashant rana

945920

1

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

1

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

|
show 3 more comments

1

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

1

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

Care to add some explanation to your undocumented answer? I didn't downvote though.

– Bazingaa
Jan 2 at 10:22

You are now accessing a global variable, not the free variable in the closure of inner

– juanpa.arrivillaga
Jan 2 at 10:23

How to do it without making la global?

– Vicky Gupta
Jan 2 at 10:26

@Bazingaa hope this is explanatory enough

– prashant rana
Jan 2 at 10:40

@juanpa.arrivillaga updated the code using the global or nonlocal way

– prashant rana
Jan 2 at 10:42

|
show 3 more comments

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 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

add a comment |

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 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

add a comment |

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 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

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 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

edited Jan 2 at 12:39

ThunderMind

639214

edited Jan 2 at 12:39

ThunderMind

639214

edited Jan 2 at 12:39

ThunderMind

639214

answered Jan 2 at 10:24

Alexander Lyapin

256

answered Jan 2 at 10:24

Alexander Lyapin

256

answered Jan 2 at 10:24

Alexander Lyapin

256

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

add a comment |

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

It is giving error: nonlocal la ^ SyntaxError: invalid syntax

– Vicky Gupta
Jan 2 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 2 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 2 at 10:32

add a comment |

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