CODEWITHSUNDEEP
pythonfunctiontry-except
kaloyan-marinov
kaloyan-marinov

Reputation: 63

How to trace back the cause of an exception raised within a function?

(This is a follow-up question to the post Python try/except: Showing the cause of the error after displaying my variables.)

I have the following script.py:

import traceback


def process_string(s):
    """
    INPUT
    -----
    s:  string

        Must be convertable to a float

    OUTPUT
    ------
    x:  float
    """

    # validate that s is convertable to a float
    try:
        x = float(s)
        return x
    except ValueError:        
        print
        traceback.print_exc()


if __name__ == '__main__':

    a = process_string('0.25')
    b = process_string('t01')
    c = process_string('201')

Upon execution of script.py, the following message is printed in the terminal window:

Traceback (most recent call last):
  File "/home/user/Desktop/script.py", line 20, in process_string
    x = float(s)
ValueError: could not convert string to float: t01

May I ask if there is a way for traceback.print_exc() to also print in the terminal window which instruction inside the if-main threw the exception which was caught by the try-except clause?

Upvotes: 5

Views: 277

Answers (1)

geckon
geckon

Reputation: 8754

What about this?

import traceback


def process_string(s):
    """
    INPUT
    -----
    s:  string

        Must be convertible to a float

    OUTPUT
    ------
    x:  float
    """
    return float(s)



if __name__ == '__main__':
    try:
        a = process_string('0.25')
        b = process_string('t01')
        c = process_string('201')
    except ValueError:        
        print
        traceback.print_exc()

Upvotes: 5

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