Python Coding Challange - Question with Answer (01280825)

 

This is a classic Python scope trap. Let’s go step by step.

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Code:

y = 50
def test():
    print(y)
    y = 20
test()

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Step 1: Look at the function test()

• Inside test, Python sees an assignment:

  y = 20

• Because of this assignment, Python treats y as a local variable for the entire function (even before it’s assigned).

• This is due to Python’s scope rules (LEGB):

  • Local (inside function)

  • Enclosed (inside outer function)

  • Global (module-level)

  • Built-in

Since y = 20 exists, Python marks y as local inside test.

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Step 2: The print(y) line

• At this point, Python tries to use the local y (because assignment makes it local).

• But the local y has not been assigned yet when print(y) runs.

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Step 3: Result

This leads to:

UnboundLocalError: local variable 'y' referenced before assignment

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✅ Answer: It raises UnboundLocalError.

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👉 If you want it to print the global y = 50, you’d need to explicitly declare it:

def test():
    global y
    print(y) 
    y = 20

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