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

UnboundLocalError Explained in Python Closures

The situation described in the question revolves around a fundamental concept in Python known as variable scope. Unlike languages with explicit variable declarations, Python determines variable scope based on assignment.

Consider the following code:

counter = 0

def increment():
  counter += 1

increment()

This code raises an UnboundLocalError. Why?

In Python, an assignment within a function marks the variable as local to that function. The line counter = 1 in the increment() function implies that counter is a local variable. However, this line attempts to access the local variable before it is assigned, leading to the UnboundLocalError.

To circumvent this issue, you have several options:

• Global Keyword: If counter is intended to be a global variable, prefix its assignment with the global keyword inside the function:

counter = 0

def increment():
  global counter
  counter += 1

• Nonlocal Keyword (Python 3.x): If counter is a local variable in a parent function and increment() is an inner function, you can use the nonlocal keyword:

def outer():
  counter = 0

  def inner():
    nonlocal counter
    counter += 1

By utilizing these techniques, you can correctly manipulate local and global variables within closures, avoiding unnecessary errors.

The above is the detailed content of Why Does `counter = 1` Inside a Python Function Cause an `UnboundLocalError`?. For more information, please follow other related articles on the PHP Chinese website!