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Python nested scopes with dynamic features

Need help with understanding the following sentence from PEP 227 and the Python Language Reference

If a variable is referenced in an enclosed scope, it is an error to delete the name. The compiler will raise a SyntaxError for 'del name'.

Lack of examples caused I couldn't reproduce an error at compile time, so an explanation with examples is highly desirable.

like image 739
trupanka Avatar asked Sep 09 '12 11:09

trupanka


1 Answers

The following raises the execption:

def foo():
    spam = 'eggs'
    def bar():
        print spam
    del spam

because the spam variable is being used in the enclosed scope of bar:

>>> def foo():
...     spam = 'eggs'
...     def bar():
...         print spam
...     del spam
... 
SyntaxError: can not delete variable 'spam' referenced in nested scope

Python detects that spam is being referenced in bar but does not assign anything to that variable, so it looks it up in the surrounding scope of foo. It is assigned there, making the del spam statement a syntax error.

This limitation was removed in Python 3.2; you are now responsible for not deleting nested variables yourself; you'll get a runtime error (NameError) instead:

>>> def foo():
...     spam = 'eggs'
...     def bar():
...         print(spam)
...     del spam
...     bar()
... 
>>> foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 6, in foo
  File "<stdin>", line 4, in bar
NameError: free variable 'spam' referenced before assignment in enclosing scope
like image 98
Martijn Pieters Avatar answered Oct 05 '22 09:10

Martijn Pieters