Python3 exception handling

Question

I'm trying to make a simple program (for learning purposes) using exception handling. What I'm trying to do is:

try: x = int(input()) except ValueError as var: # HERE I WANT TO USE THE 'x' STRING VALUE

I know about the the various ways to use the exception message (for instance by using str(var)).

For example, if the input is bla which would cause a ValueError exception, print(var) outputs invalid literal for int() with base 10: 'bla', which is not really usable.

However, I created this program that should use str(var) to my advantage:

try: x = int(input()) except ValueError as var: s = str(var) print('Exception message: ' +s) i=0 while (s[i] != '\''): i = i+1 i = i+1 print_str = '' while (s[i] != '\''): print_str = print_str + str(s[i]) i++ print('print_str = ' + print_str)

This program would probably work after a few changes..

So my question is: Is there a more direct way to get the 'x' string value?

Yes, i++ is invalid syntax. It has nothing to do with exception handling. — Lev Levitsky
– Lev Levitsky, Commented Jun 10, 2014 at 20:59
Fixed it, my bad(probably because I use C++ more often). That lower program was not the main purpose of this question anyway.. — dantc12
– dantc12, Commented Jun 10, 2014 at 21:03

jonrsharpe · Accepted Answer · 2014-06-10 21:00:04Z

You can't access x in the except because it was never assigned - the exception is thrown somewhere in int(input(...)), before x = happens. Instead, do:

x = input(...) try: x = int(x) except ValueError: print("Couldn't make '{0}' an integer.".format(x))

Community · Accepted Answer · 2017-05-23 11:43:24Z

Furthering jonsharpe's response, you cannot catch most SyntaxErrors as they happen at compile time. Thus there's no hope whatsoever of even running the code, never mind recovering from the error.

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