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Home/ Questions/Q 8516543
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T05:27:31+00:00 2026-06-11T05:27:31+00:00

Possible Duplicate: Short Description of Python Scoping Rules I wrote two simple functions: #

  • 0

Possible Duplicate:
Short Description of Python Scoping Rules

I wrote two simple functions:

# coding: utf-8
def test():
    var = 1 
    def print_var():
        print var 
    print_var()
    print var 

test()
# 1
# 1
def test1():
    var = 2 
    def print_var():
        print var 
        var = 3 
    print_var()
    print var 

test1()
# raise Exception

In comparison, test1() assigns value after print var, then raise an Exception: UnboundLocalError: local variable 'var' referenced before assignment, I think the moment I call inner print var, var has a value of 2, am I wrong?

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  1. Editorial Team
    Editorial Team
    2026-06-11T05:27:32+00:00Added an answer on June 11, 2026 at 5:27 am

    Yes, you’re incorrect here. Function definition introduces a new scope.

    # coding: utf-8
    def test():
        var = 1 
        def print_var():
            print var    <--- var is not in local scope, the var from outer scope gets used
        print_var()
        print var 
    
    test()
    # 1
    # 1
    def test1():
        var = 2 
        def print_var():
            print var     <---- var is in local scope, but not defined yet, ouch
            var = 3 
        print_var()
        print var 
    
    test1()
    # raise Exception
    
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