Why Does Python Raise an UnboundLocalError, and How Can I Fix It?

Understanding Python's UnboundLocalError: Resolving Variable Scope Issues

When assigning values to variables within functions, Python can raise an UnboundLocalError if a variable is accessed before being defined locally within the function. Let's delve into the reasons behind this error and explore solutions.

One common scenario leading to this error is when a variable is declared with an assignment statement within a function, effectively shadowing a global variable of the same name. Consider the following code snippet:

Var1 = 1
Var2 = 0
def function():
    if Var2 == 0 and Var1 > 0:
        print("Result 1")
    elif Var2 == 1 and Var1 > 0:
        print("Result 2")
    elif Var1 < 1:
        print("Result 3")
    Var1 -= 1
function()

In this example, the function attempts to use the variable Var1, which has already been declared in the global scope. However, the line Var1 -= 1 within the function assigns a new value to a local variable Var1, overshadowing the global variable. When Python encounters this, it raises an UnboundLocalError.

Resolving the Error

To fix this issue, one approach is to use Python's global keyword within the function. This keyword explicitly informs Python that the variable referenced within the function is the global one, not a distinct local variable.

def function():
    global Var1, Var2  # Declare global variables inside function
    if Var2 == 0 and Var1 > 0:
        print("Result 1")
    elif Var2 == 1 and Var1 > 0:
        print("Result 2")
    elif Var1 < 1:
        print("Result 3")
    Var1 -= 1

By using the global keyword, the function can access and modify the global variables Var1 and Var2.

Considerations:

While using global variables can be tempting, it's generally discouraged in Python as it can lead to code that is difficult to maintain and debug. Instead, it's preferable to pass variables as arguments to functions or utilize class variables when necessary.

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