Why Does My Python 3 Code Throw an `UnboundLocalError`?

Unresolved Variable Reference: Understanding the 'UnboundLocalError' in Python 3

The code snippet provided has encountered the 'UnboundLocalError' exception, indicating that it references a local variable ('Var1') before assigning it a value. This occurs when the variable name is declared within a function, but no value is assigned to it prior to its use.

To rectify this error, we delve deeper into the nuances of variable scoping in Python 3. While the code initially declares 'Var1' as a global variable, the assignment statement 'Var1 -= 1' within the function creates a local variable called 'Var1'. This modifies the variable local to the function, not the global one declared outside.

To utilize global variables within a function, Python 3 provides the 'global' keyword. By adding 'global Var1, Var2' at the beginning of the function, we explicitly state that we intend to reference the global variables named 'Var1' and 'Var2' within the function's scope. This prevents the creation of local variables with the same names, thus resolving the 'UnboundLocalError'.

In Python 3, the 'nonlocal' statement also exists for manipulating variables defined in an enclosing scope. However, 'nonlocal' is typically used when nesting functions, which is not the case in the provided code.

For further insights into variable scoping and error handling, the Python website and related documentation serve as valuable resources. By understanding the principles of variable referencing and leveraging the appropriate keywords, Python developers can effectively manage variable scope and avoid such runtime errors.

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