UnboundLocalError: local variable referenced before assignment

The error message:

UnboundLocalError: local variable 'variable_name' referenced before assignment

occurs when you try to use a local variable inside a function before it has been assigned a value.

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1. What Causes This Error?

This error mainly happens when Python assumes a variable inside a function is local, but it is not assigned before being used.

Example of the Error:

x = 10  # Global variable

def modify():
    print(x)  # Trying to use 'x' before assigning a value
    x = x + 5  # Local assignment

modify()

Error Output:

UnboundLocalError: local variable 'x' referenced before assignment

Why Does This Happen?

• Even though x is defined globally, Python treats x inside modify() as a local variable because it is being assigned (x = x + 5).
• Since Python processes the function’s entire scope before execution, it marks x as a local variable.
• The function tries to access x before it is assigned, leading to UnboundLocalError.

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2. Solutions to Fix the Error

Solution 1: Use the global Keyword for Global Variables

If you want to modify a global variable inside a function, use the global keyword.

Corrected Code:

x = 10  # Global variable

def modify():
    global x  # Tell Python to use the global 'x'
    print(x)  # No error
    x = x + 5  # Modify global 'x'

modify()
print(x)  # Output: 15

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Solution 2: Initialize the Variable Before Using It

If you don’t want to modify a global variable, ensure the variable is assigned before using it inside the function.

Incorrect Code:

def example():
    print(y)  # Error: 'y' is used before assignment
    y = 20  # Local assignment

example()

Corrected Code:

def example():
    y = 20  # Define before using
    print(y)  # No error

example()

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Solution 3: Use nonlocal for Variables Inside Nested Functions

If a variable is inside an outer function, but you need to modify it inside an inner function, use nonlocal.

Incorrect Code:

def outer():
    x = 5  # Local to 'outer'

    def inner():
        x = x + 1  # Error: Python thinks 'x' is local
        print(x)

    inner()

outer()

Corrected Code:

def outer():
    x = 5  # Local to 'outer'

    def inner():
        nonlocal x  # Use 'x' from 'outer'
        x = x + 1
        print(x)

    inner()

outer()  # Output: 6

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Solution 4: Use Default Arguments to Avoid Scope Issues

You can pass the variable as a function parameter to avoid scope confusion.

Corrected Code:

x = 10

def modify(x):  # Pass 'x' as a parameter
    x = x + 5
    print(x)

modify(x)  # Output: 15
print(x)  # Global 'x' is still 10