Why Does `counter = 1` Inside a Function Cause an `UnboundLocalError` in Python?

UnboundLocalError in Closures: An Analysis

An UnboundLocalError occurs when a variable used within a function's code is not defined locally, globally, or as part of a nonlocal block. Consider the following code:

counter = 0

def increment():
  counter += 1

increment()

Running this code will result in an UnboundLocalError. Why does this occur?

Understanding Python's Variable Scope

Python dynamically determines variable scope based on assignment. If a variable is assigned within a function, it is considered local to that function. In our example, the assignment counter = 1 within increment() implicitly defines counter as local to that function.

Local vs. Global Variables

Python distinguishes between local and global variables. Global variables are declared at the module level and are accessible throughout the program. Local variables, on the other hand, are created within functions and only exist within those functions.

In our case, counter is not defined globally. The error occurs because Python tries to read the value of counter from the local scope of increment() before it has been assigned, hence the UnboundLocalError.

Resolving the Error

To resolve this error, you can do one of the following:

• Use the global Keyword: For variables that should be accessible both within the function and outside of it, use the global keyword. This declares the variable as global, allowing it to be accessed from any scope.
• Use nonlocal (Python 3.x only): This keyword is used to declare a variable as nonlocal, meaning that it is not local to the current function but is accessible within its enclosing non-nested functions.

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