If you assign a value to a variable anywhere inside a function, Python assumes that variable is local to that function. So even if there\u2019s a global variable with the same name, Python won\u2019t look at it. It treats your variable as a separate, uninitialized local one. That\u2019s why you get the error when you try to use it before assigning a value.<\/p>\n\n\n\n
Want to fix it? You\u2019ve got two options:<\/strong><\/p>\n\n\n\n You’re using a global variable inside a function, but Python throws this error:<\/strong><\/p>\n\n\n\n Output:<\/strong><\/p>\n\n\n\n You’re confused \u2014 counter is clearly defined at the top level. Why is Python treating it like it\u2019s undefined inside the function?<\/p>\n\n\n\n This error occurs when you try to modify a global variable without declaring it as global inside a function. In Python, when you assign a value to a variable anywhere inside a function, Python treats it as a local variable unless explicitly told otherwise.<\/p>\n\n\n\n So in this line:<\/strong><\/p>\n\n\n\n Python treats counter as a new local variable. But since it hasn’t been assigned before that line, it throws an UnboundLocalError.<\/p>\n\n\n\n Even though counter exists globally, Python doesn’t look at that definition because it assumes you’re trying to create a new local variable.<\/p>\n\n\n\n To tell Python that you’re referring to the global counter, use the global keyword:<\/p>\n\n\n\n Now Python knows you’re not creating a new local variable \u2014 you’re modifying the global one.<\/p>\n\n\n\n Using global can make code harder to debug and maintain. Instead, consider returning values and reassigning them:<\/p>\n\n\n\n This makes data flow clear and avoids side effects.<\/p>\n\n\n\n If you’re working with nested functions<\/strong> and want to modify a variable from the enclosing scope (but not global), use nonlocal:<\/p>\n\n\n\n This modifies counter in the outer() function, not the global scope.<\/p>\n\n\n\n Best Practices:<\/strong><\/p>\n\n\n\n Understanding Python\u2019s scoping rules\u2014Local, Enclosing, Global, and Built-in (LEGB)\u2014is key to avoiding bugs like UnboundLocalError.<\/p>\n\n\n\n\n
Description of the Problem<\/h2>\n\n\n\n
counter = 10\n\ndef increment():\n counter += 1\n print(counter)\n\nincrement()<\/code><\/pre>\n\n\n\nUnboundLocalError: local variable 'counter' referenced before assignment<\/code><\/pre>\n\n\n\nWhy This Happens<\/h2>\n\n\n\n
counter += 1<\/code><\/pre>\n\n\n\nHow to Resolve the UnboundLocalError in 3 Simple Steps<\/h2>\n\n\n\n
Step 1: Use the global keyword if you want to modify the global variable<\/h3>\n\n\n\n
counter = 10\n\ndef increment():\n global counter\n counter += 1\n print(counter)\n\nincrement() # Output: 11<\/code><\/pre>\n\n\n\nStep 2: Avoid global when possible – prefer return values<\/h3>\n\n\n\n
def increment(counter):\n counter += 1\n return counter\n\ncounter = 10\ncounter = increment(counter)\nprint(counter) # Output: 11<\/code><\/pre>\n\n\n\nStep 3: Use nonlocal for modifying variables in nested functions<\/h3>\n\n\n\n
def outer():\n counter = 0\n\n def inner():\n nonlocal counter\n counter += 1\n print(counter)\n\n inner()\n inner()\n\nouter()<\/code><\/pre>\n\n\n\n\n
Conclusion<\/h2>\n\n\n\n