Why Does Python Throw an UnboundLocalError?

How UnboundLocalError Occurs: Unbound Names and Variable Binding in Python

In Python, variable binding determines the scope and lifetime of variables. When a name is not assigned a value, it is considered unbound. This can lead to the UnboundLocalError exception.

Understanding Unbound Local Variables

Unlike languages with explicit declarations, Python allows variable assignment anywhere within a block. However, if a name in a function is used before it is assigned, an UnboundLocalError is raised. This occurs because the compiler cannot determine the variable's value since it hasn't been bound yet.

Example: Code That Causes UnboundLocalError

Consider the following code:

def foo():
    if False:
        spam = 'eggs'
    print(spam)

foo()

This code results in an UnboundLocalError because the spam variable is being used in the print statement without first being assigned. Even though the if statement checks a condition, it doesn't execute the assignment, leaving spam unbound.

Binding Operations in Python

Variables become bound through various operations:

• Assignment
• for loop iteration
• Function parameters
• Import statements
• Caught exceptions (except clause)
• Context managers (with statement)

When a name is bound within a scope, such as a function, it becomes a local variable. However, using the global (or nonlocal in Python 3) statement explicitly declares a name as global, allowing it to be referenced and modified from outside the scope.

Preventing UnboundLocalError

To avoid UnboundLocalError, ensure that variables are bound properly before they are used. This can be done by:

• Assigning values to variables before using them.
• Using the global statement to declare global variables within functions.
• Avoiding unreachable code blocks.

References:

• [Python Reference Documentation - Naming and Binding](https://docs.python.org/3/reference/introduction.html#naming-and-binding)

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