Programming Forums - Why did %S break my if statement?

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Why did %S break my if statement?

I was working on an if statement inside a function (see below for code) and made the small typo of %S instead of %s inside of one of my print statements. For some reason this caused the program to exit the function immediately. Can someone explain exactly what that is doing?

I don't know that it matters, but here is the function with the location of the (now former) typo in red. The gratuitous print statments that I used to help me debug are still in there, so you can ignore them.

def remove_person():

    name = input_list()

    if dbase.has_key(name):

            print 'dbase has %s' % (name)

        print len(dbase[name])

        if not len(dbase[name]) == 0:

            print "%s still has items on loan." % (name)

            certain = raw_input("Delete anyway? (y/N):")

            if re.match('^[yY][eE]?[sS]?$', certain):

                del dbase[name]

                print 'deleted'

            else:

                print "not deleted"

            print 'delete path NOT taken'

        else:

            print 'delete path taken'

            del dbase[name]

    else:

        print "%s was not found." % (name)

What do you mean it "exited the function"? It should be raising a ValueError, since %S is an unsupported format character. Is this Python 2.4?

!!! It makes so much sense now. It was rasing a ValueError, but that function was run from inside a try block. just before the function is called, I'd had the program try int(some.user.input) and had a ValueError exception ready in case of bad input.

That function was called at option[choice]() below.

while choice != 9:        # main menu loop

    menu()

    try:

        choice = int(raw_input("Selection:"))

        if choice != 9:

            try:      

                option[choice]()



            except KeyError:

                print "Invalid choice: please enter a number from the menu"



    except ValueError:

        print "Invalid choice: please enter a number"

print ""

Thanks!

Ahh. Excellent. What a sucker those things can be. My first thought was you had it in a try, since you didn't mention raising any errors. But I couldn't see that anywhere in your code. Very well. :P