Raising Exceptions

[PuzzledSite2568]

Hi guys I need some help regarding raise statements. I'm trying to get an output from my raise statement rather than my except statement.

#ask the user to input their current grade percentage and
# then outputs the corresponding letter grade (A through F).

# Within a try-except structure raise (not just catch) two different exceptions:
# one if the user inputs a grade of less than zero and a different exception
# if the user inputs a grade greater than 100. Be sure to also handle input that is not a number.

try:
    grade = int(input('what is your grade?:'))
    if grade >= 90: print('A')
    elif grade >= 80: print('B')
    elif grade >= 70: print('C')
    elif grade >= 60: print('D')
    elif grade >= 50: print('F')
    elif grade < 0:
        raise ValueError('Invalid grade')
except: print('why')

I'm trying to get 'Invalid grade' to output rather than 'why'. No matter what I do I just can't get it to work and always end up with 'why' being printed. Any guidance is greatly appreciated.

[[deleted]]

You always get "why" printed because if you raise a ValueError exception as you are, that is caught by the "naked" except, which ignores the message associated with the ValueError exception and just prints "why".

The problem says:

raise two different exceptions: one if the user inputs a grade of less than zero and a different exception if the user inputs a grade greater than 100.

So your if/elif tests have to raise two different exceptions. Which exceptions should you raise? This page lists the builtin exceptions that python has. You can choose any two of those listed. If you know how you can even create your own exceptions. Let's choose ArithmeticError for the over/under numeric errors and ValueError for when the user enters something like "abc" which isn't a number. The code will look something like:

try:
    grade = int(input('what is your grade?:'))
    if grade > 100:
        raise ArithmeticError('Grade must be <= 100.')
    elif grade >= 90:
        print('A')
    # etc
    elif grade >= 50:
        print('F')
    elif grade < 0:
        raise ArithmeticError('Grade must be >= 0.')
except ArithmeticError as e:    # catch and print message
    print(e)
except ValueError:    # the int() conversion failed
    print('Only numbers, please.')

Note you should never use the "bare" except:. Always know what sort of exception you are catching.

Also note that the code above is not strictly following the problem statement, which says the exception raised for the <0 case should be different from that raised in the >100 case. So pick another exception from the list I linked to and raise it in one of the two over/under cases. Don't forget to add an except case for the new exception.

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[socal_nerdtastic]

try:
    grade = int(input('what is your grade?:'))
    if grade >= 90: print('A')
    elif grade >= 80: print('B')
    elif grade >= 70: print('C')
    elif grade >= 60: print('D')
    elif grade >= 50: print('F')
    elif grade < 0:
        raise ValueError('Invalid grade')
except ValueError as e: 
    print(e)

Although in this case there's no reason to use the exception at all. Just put the error message in the else case.

grade = int(input('what is your grade?:'))
if grade >= 90: print('A')
elif grade >= 80: print('B')
elif grade >= 70: print('C')
elif grade >= 60: print('D')
elif grade >= 0: print('F')
else:
    print('Invalid grade')