Programming Forums > Script Programming > Python

somehollis · Jun 22nd, 2006, 8:26 PM

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.

(Toggle Plain Text)

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)

Sane · Jun 22nd, 2006, 8:35 PM

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?

somehollis · Jun 22nd, 2006, 8:46 PM

!!! 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.

(Toggle Plain Text)

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!

Sane · Jun 22nd, 2006, 8:49 PM

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.

Jun 22nd, 2006, 8:26 PM	#1
somehollis Programmer Join Date: May 2006 Location: Memphis, TN Posts: 31 Rep Power: 0	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. (Toggle Plain Text) 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]?$', certain): del dbase[name] print 'deleted' else: print "not deleted" print 'delete path NOT taken' else: print 'delete path taken' del dbase[name] else: print "%s was not found." % (name) def remove_person(): name = input_list() if dbase.has_key(name): print 'dbase has %s' % (name) print len(dbase[name]) if not len(dbase[name]) == 0: print "%s still has items on loan." % (name) certain = raw_input("Delete anyway? (y/N):") if re.match('^[yY][eE]?[sS]?$', 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)

Jun 22nd, 2006, 8:46 PM	#3
somehollis Programmer Join Date: May 2006 Location: Memphis, TN Posts: 31 Rep Power: 0	!!! 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. (Toggle Plain Text) 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 "" 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!

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Jun 22nd, 2006, 8:35 PM	#2
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,869 Rep Power: 5	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?

Jun 22nd, 2006, 8:49 PM	#4
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,869 Rep Power: 5	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.

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Programming Forums > Script Programming > Python

Why did %S break my if statement?