CODEWITHSUNDEEP
python
John Howard
John Howard

Reputation: 64125

Changing variable leads to referenced before assignment

I am experiencing by far the weirdest error I've come across right now. My code is simple:

wtf = 5
def update():
    print(wtf)
    wtf = 1

update()

With the line # wtf = 1 commented out, all works fine, and 5 is printed. However, if I uncomment out # wtf = 1, before I can even print out wtf (5), I get an UnboundLocalError: local variable 'wtf' referenced before assignment. I have no idea WTF is going on here. Why is this happening?

Upvotes: 1

Views: 137

Answers (2)

Ry-
Ry-

Reputation: 224867

When there’s potential for a variable to be assigned to in a scope, that variable becomes local to that scope. You can prevent that behaviour using global for global variables, or nonlocal for any non-local in Python 3:

wtf = 5
def update(dt):
    nonlocal wtf
    print(wtf)
    wtf = 1

Upvotes: 1

TheSoundDefense
TheSoundDefense

Reputation: 6935

When all you have is print(wtf) in that function, Python assumes that you're just trying to print the global wtf. If you add in wtf = 1, then Python is forced to assume you're trying to change a local variable - you can't assign new values to a global variable in a function, unless you use global wtf at the top. So in that second case, Python assumes that wtf is a local, which is why the print(wtf) statement fails - you're now trying to print a local variable before it was assigned, or so Python thinks.

To fix this, add global wtf as the first line of your update function.

Upvotes: 2

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