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Home/ Questions/Q 9173279
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T16:33:09+00:00 2026-06-17T16:33:09+00:00

I am learning Python. A book on Python 3 says the following code should

  • 0

I am learning Python. A book on Python 3 says the following code should work fine:

def funky():
    print(myvar)
    myvar = 20
    print(myvar)

myvar = 10
funky()

But when I run it in Python 3.3, I got the

UnboundLocalError: local variable 'myvar' referenced before assignment

error. My understanding is that the first print(myvar) in funky should be 10 since it’s a global variable. The second print(myvar) should be 20 since a local myvar is defined to be 20. What is going on here? Please help clarify.

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  1. Editorial Team
    Editorial Team
    2026-06-17T16:33:10+00:00Added an answer on June 17, 2026 at 4:33 pm

    You need to call global in your function before assigning a value.

    def funky():
        global myvar
        print(myvar)
        myvar = 20
        print(myvar)
    
    myvar = 10
    funky()
    

    Note that you can print the value without calling global because you can access global variables without using global, but attempting to assign a value will require it.

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