gCharlie = 0
gJeff = 0
def Bob ():
Charlie = gCharlie
Jeff = gJeff
Number = int(input("Hello and Welcome to Charlie's Number Guessing Game. Enter a nonnegative number from 0 to 10 not counting 0. "))
This code gives me this error in Python 3.2:
UnboundLocalError: local variable 'gCharlie' referenced before assignment
What does this local variable error mean?
inside the Scope of your function you must have reassigned gJeff and gCharlie, which created them as new local variables. To tell python that you're using the globals, change the top of your function to look like this.
def Bob():
global gCharlie
global gJeff
Charlie=gCharlie
without telling python that you're using the globals, it tries to reference local gCharlie and gJeff variables, which as it says, have not been assigned at that point in your function. The reason people are getting it to work is because they're using only the code you've posted. You can reference globals without explicitly saying so ONLY if you don't change their values within the function referencing them.
As a rule python searches in this order for a variable name: local scope, any def it is nested inside of, global, built ins. Bbecause gJeff and gCharlie are local variables in your function, it stops there, unless you tell it otherwise.
If you want to see this in action try to look at this
x=5
def useX():
x=0 #this is a local variable
print "in function: ", x
def main():
print "in main(1): ", x
useX()
print "in main(2): ", x
main()
this will output
in main(1): 5
in function: 0
in main(2): 5
because within the function, x is created as a new local variable. Adding a global x statement to the useX function would change the last line to print "0" instead of "5"
import some_module at a module-level is an ordinary global that you obviously can use inside any function. You don't need global if you don't change their bindings inside the function, example. If you do need it then you should probably refactor the function into a method and the global into an instance attribute.global useX to call useX() inside main().
It might be due to there is gCharlie = inside a function (note: the first letter is g).
Use parameters instead of globals:
def bob(charlie=0, jeff=0):
number = int(input("..."))
# ...
bob(charlie=3)
It means that you're assigning to gCharlie in the part of the function you didn't show, and so the Python compiler has marked it as a local variable. As such, you're accessing it before it exists. Use nonlocal or global to solve.
Two previous answers here are correct, but both are a bit unclear. I'll show with some examples:
The code you show will work fine:
>>> gCharlie = "Global!"
>>>
>>> def foo():
... print(gCharlie)
...
>>> foo()
Global!
>>> print(gCharlie)
Global!
So that's not the problem at all. However, you can't assign global variables inside a function:
>>> gCharlie = "Global!"
>>> def foo():
... gCharlie = "Local!"
... print(gCharlie)
...
>>> foo()
Local!
>>> print(gCharlie)
Global!
As you see, the global variable gCharlie did not change. This is because you did not modify it, you created a new local variable, with the same name. And this is the cause of the error:
>>> gCharlie = "Global!"
>>> def foo():
... oldCharlie = gCharlie
... gCharlie = "Local!"
... print(gCharlie)
...
>>> foo()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in foo
UnboundLocalError: local variable 'gCharlie' referenced before assignment
The hint is in the error. It says local variable gCharlie. The problem is not the inability to access the global gCharlie, but that the local one hasn't been created yet.
The fix is to specify that you don't want to create a local variable, but that you want to modify the global one. You do this with the global keyword.
>>> gCharlie = "Global!"
>>> def foo():
... global gCharlie
... oldCharlie = gCharlie
... gCharlie = "Local!"
... print(gCharlie)
...
>>> foo()
Local!
>>> print(gCharlie)
Local!
As you see now, you modified the global variable.
That said, global variables are usually a bad idea. Avoid them. Try to pass in the variables as parameters instead.
