Python Coding Challange - Question with Answer (01150925)

 

Step 1: Global Variable

x = 100

Here, a global variable x is created with value 100.

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Step 2: Inside test()

def test():
    print(x) 
    x = 50

• Python sees the line x = 50 inside the function.

• Because of this, Python treats x as a local variable within test().

• Even though the print(x) comes before x = 50, Python already marks x as a local variable during compilation.

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Step 3: Execution

• When print(x) runs, Python tries to print the local x.

• But local x is not yet assigned a value (since x = 50 comes later).

• This causes an UnboundLocalError.

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Error Message

UnboundLocalError: local variable 'x' referenced before assignment

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✅ In simple words:
Even though x = 100 exists globally, the function test() creates a local x (because of the assignment x = 50).
When you try to print x before assigning it, Python complains.

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👉 If you want to fix it and use the global x, you can do:

x = 100
def test():
    global x
    print(x)
    x = 50
test()

This will print 100 and then change global x to 50.

BIOMEDICAL DATA ANALYSIS WITH PYTHON