How to force Python to report an error if a local variable is referenced out of its scope?

Question

list1 = ["AAA", "BBB"]
for item in list1:
    print(item)

print (item) # <--- out of scope, but Python doesn't report any error

For the code above, although item is out of its scope, Python will not report an error.

Is it possible to force Python to report an error?

`for` loops don't create any scope, so there is no error to report here. See https://stackoverflow.com/questions/291978/short-description-of-the-scoping-rules — Thierry Lathuille, Apr 26 '21 at 15:34
@ThierryLathuille no error except for "unexpected token '<'", etc. — Nicholas Hunter, Apr 26 '21 at 15:36
No, but what would you want that? What is the actual problem that you encounter? — Thierry Lathuille, Apr 26 '21 at 15:42
The immediate solution that comes to mind is wrap in a function, but as Thierry said - why do you need that behavior? — Tomerikoo, Apr 26 '21 at 15:54
I think this is an error prone issue, especially for the beginner of python. Someone ask me to debug the issue of their code, and I found this is due to the local variable scope — camino, Apr 26 '21 at 17:13

score 0 · Accepted Answer · answered Apr 26 '21 at 17:13

the variable that's being used in a loop will eventually arrive at the index "-1" of an iterable object. So each time you use that same variable used in a previous loop, it will return list1[-1] which is indeed the last element in every iterable object in python

Solution: you can use del keyword to delete that variable

list1 = ["AAA", "BBB"]
for item in list1:
    print(item)
del item    #now item is not a defined variable in our program.
print (item) #<--- will throw an error because the variable "item" no longer exists

NameError will be raised

score -1 · Answer 2 · answered Apr 26 '21 at 15:50

-1

Functions have limited scope. So keep for loop inside function

def f():
  for item in list1:
    print(item)
print(item)

It will throw an error but you need to call a function whenever it required.

answered Apr 26 '21 at 15:50

Zalak Bhalani

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How to force Python to report an error if a local variable is referenced out of its scope?

2 Answers2