I Expected this code to throw an error as ival(local variable) is used outside the scope of try except, but why is it running without any error

Asked Jul 24 '19 at 07:12

Active Jul 24 '19 at 07:23

Viewed 94 times

I expected the following code to throw an error as the variable ival(local variable) is being used outside of its scope i.e., in if condition. I didn't encounter any issues/errors why? Please explain the reason.

rawstr=input('Enter a number')
try:
    ival=int(rawstr)
except:
    ival=-1
if ival>0:
    print('Nice work')
else:
    print('Not a Number')

I expected a traceback error saying ival is not defined.

edited Jul 24 '19 at 07:23

quamrana

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asked Jul 24 '19 at 07:12

Gopi B

`try` does not introduce scope. `ival` is just a global like `rawstr`. – quamrana Jul 24 '19 at 07:22
The variable `ival` is local for the function, not only for the `try` part. See here: https://stackoverflow.com/questions/291978/short-description-of-the-scoping-rules – ZiGaelle Jul 24 '19 at 07:26

I Expected this code to throw an error as ival(local variable) is used outside the scope of try except, but why is it running without any error

0 Answers0