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UnboundLocalError Local variable Referenced Before Assignment in Python

Handling errors is an integral part of writing robust and reliable Python code. One common stumbling block that developers often encounter is the “UnboundLocalError” raised within a try-except block. This error can be perplexing for those unfamiliar with its nuances but fear not – in this article, we will delve into the intricacies of the UnboundLocalError and provide a comprehensive guide on how to effectively use try-except statements to resolve it.

What is UnboundLocalError Local variable Referenced Before Assignment in Python?

The UnboundLocalError occurs when a local variable is referenced before it has been assigned a value within a function or method. This error typically surfaces when utilizing try-except blocks to handle exceptions, creating a puzzle for developers trying to comprehend its origins and find a solution.

Why does UnboundLocalError: Local variable Referenced Before Assignment Occur?

below, are the reasons of occurring “Unboundlocalerror: Try Except Statements” in Python :

Variable Assignment Inside Try Block

Reassigning a global variable inside except block.

• Accessing a Variable Defined Inside an If Block

In the below code, example_function attempts to execute some_operation within a try-except block. If an exception occurs, it prints an error message. However, if no exception occurs, it prints the value of the variable result outside the try block, leading to an UnboundLocalError since result might not be defined if an exception was caught.

In below code , modify_global function attempts to increment the global variable global_var within a try block, but it raises an UnboundLocalError. This error occurs because the function treats global_var as a local variable due to the assignment operation within the try block.

Solution for UnboundLocalError Local variable Referenced Before Assignment

Below, are the approaches to solve “Unboundlocalerror: Try Except Statements”.

Initialize Variables Outside the Try Block

Avoid reassignment of global variables.

In modification to the example_function is correct. Initializing the variable result before the try block ensures that it exists even if an exception occurs within the try block. This helps prevent UnboundLocalError when trying to access result in the print statement outside the try block.

Below, code calculates a new value ( local_var ) based on the global variable and then prints both the local and global variables separately. It demonstrates that the global variable is accessed directly without being reassigned within the function.

In conclusion , To fix “UnboundLocalError” related to try-except statements, ensure that variables used within the try block are initialized before the try block starts. This can be achieved by declaring the variables with default values or assigning them None outside the try block. Additionally, when modifying global variables within a try block, use the `global` keyword to explicitly declare them.

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[SOLVED] Local Variable Referenced Before Assignment

Python treats variables referenced only inside a function as global variables. Any variable assigned to a function’s body is assumed to be a local variable unless explicitly declared as global.

Why Does This Error Occur?

Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.

Before we hop into the solutions, let’s have a look at what is the global and local variables.

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

The variable myVar has been assigned a value twice. Once before the declaration of myFunction and within myFunction itself.

Using Global Variables

Passing the variable as global allows the function to recognize the variable outside the function.

Create Functions that Take in Parameters

Instead of initializing myVar as a global or local variable, it can be passed to the function as a parameter. This removes the need to create a variable in memory.

UnboundLocalError: local variable ‘DISTRO_NAME’

This error may occur when trying to launch the Anaconda Navigator in Linux Systems.

Upon launching Anaconda Navigator, the opening screen freezes and doesn’t proceed to load.

Try and update your Anaconda Navigator with the following command.

If solution one doesn’t work, you have to edit a file located at

After finding and opening the Python file, make the following changes:

In the function on line 159, simply add the line:

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

The program takes information from a form filled out by a user. Accordingly, an email is sent using the information.

Upon running you get the following error:

We have created a class myForm that creates instances of Django forms. It extracts the user’s name, email, and message to be sent.

A function GetContact is created to use the information from the Django form and produce an email. It takes one request parameter. Prior to sending the email, the function verifies the validity of the form. Upon True , .get() function is passed to fetch the name, email, and message. Finally, the email sent via the send_mail function

Why does the error occur?

We are initializing form under the if request.method == “POST” condition statement. Using the GET request, our variable form doesn’t get defined.

Local variable Referenced before assignment but it is global

This is a common error that happens when we don’t provide a value to a variable and reference it. This can happen with local variables. Global variables can’t be assigned.

This error message is raised when a variable is referenced before it has been assigned a value within the local scope of a function, even though it is a global variable.

Here’s an example to help illustrate the problem:

In this example, x is a global variable that is defined outside of the function my_func(). However, when we try to print the value of x inside the function, we get a UnboundLocalError with the message “local variable ‘x’ referenced before assignment”.

This is because the += operator implicitly creates a local variable within the function’s scope, which shadows the global variable of the same name. Since we’re trying to access the value of x before it’s been assigned a value within the local scope, the interpreter raises an error.

To fix this, you can use the global keyword to explicitly refer to the global variable within the function’s scope:

However, in the above example, the global keyword tells Python that we want to modify the value of the global variable x, rather than creating a new local variable. This allows us to access and modify the global variable within the function’s scope, without causing any errors.

Local variable ‘version’ referenced before assignment ubuntu-drivers

This error occurs with Ubuntu version drivers. To solve this error, you can re-specify the version information and give a split as 2 –

Here, p_name means package name.

With the help of the threading module, you can avoid using global variables in multi-threading. Make sure you lock and release your threads correctly to avoid the race condition.

When a variable that is created locally is called before assigning, it results in Unbound Local Error in Python. The interpreter can’t track the variable.

Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.

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Fixing Python UnboundLocalError: Local Variable ‘x’ Accessed Before Assignment

Understanding unboundlocalerror.

The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope. This error typically surfaces when using a variable that has not been initialized in the current function’s scope or when an attempt is made to modify a global variable without proper declaration.

Solutions for the Problem

To fix an UnboundLocalError, you need to identify the scope of the problematic variable and ensure it is correctly used within that scope.

Method 1: Initializing the Variable

Make sure to initialize the variable within the function before using it. This is often the simplest fix.

Method 2: Using Global Variables

If you intend to use a global variable and modify its value within a function, you must declare it as global before you use it.

Method 3: Using Nonlocal Variables

If the variable is defined in an outer function and you want to modify it within a nested function, use the nonlocal keyword.

That’s it. Happy coding!

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How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python

You could also see this error when you forget to pass the variable as an argument to your function.

How to reproduce this error

How to fix this error.

I hope this tutorial is useful. See you in other tutorials.

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Local variable referenced before assignment in Python

Last updated: Apr 8, 2024 Reading time · 4 min

# Local variable referenced before assignment in Python

The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.

To solve the error, mark the variable as global in the function definition, e.g. global my_var .

Here is an example of how the error occurs.

We assign a value to the name variable in the function.

# Mark the variable as global to solve the error

To solve the error, mark the variable as global in your function definition.

If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .

# Local variables shadow global ones with the same name

You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.

Accessing the name variable in the function is perfectly fine.

On the other hand, variables declared in a function cannot be accessed from the global scope.

The name variable is declared in the function, so trying to access it from outside causes an error.

Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.

# Returning a value from the function instead

An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.

We simply return the value that we eventually use to assign to the name global variable.

# Passing the global variable as an argument to the function

You should also consider passing the global variable as an argument to the function.

We passed the name global variable as an argument to the function.

If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global .

# Assigning a value to a local variable from an outer scope

If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.

The nonlocal keyword allows us to work with the local variables of enclosing functions.

Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.

Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.

Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.

Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.

# Discussion

As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:

• Becomes local to the scope.
• Shadows any variables from the outer scope that have the same name.

The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.

At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.

The most intuitive way to solve the error is to use the global keyword.

The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.

• If a variable is only referenced inside a function, it is implicitly global.
• If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .

If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• SyntaxError: name 'X' is used prior to global declaration

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

Decode Python

Python Tutorials & Tips

Fixing ‘UnboundLocalError’ in Python: A Simple Guide with Code Samples

Python is a popular programming language that is widely used for developing various applications. However, like any other programming language, it is not free from errors. One of the common errors that Python developers encounter is the ‘UnboundLocalError’. This error occurs when a local variable is referenced before it is assigned a value. In this article, we will discuss in detail what ‘UnboundLocalError’ is, why it occurs, and how to fix it.

When a variable is defined inside a function, it is considered a local variable. If the function tries to access this variable before it is assigned a value, it results in an ‘UnboundLocalError’. This error can be frustrating for developers, especially when they are working on a large project. However, it is not difficult to fix this error. One of the ways to fix it is by using the ‘global’ keyword to declare the variable as a global variable.

In conclusion, understanding ‘UnboundLocalError’ in Python is crucial for developers who want to avoid errors in their code. By following best practices and using the right techniques, developers can easily fix this error and ensure that their code runs smoothly. In the next section, we will explore in detail how to fix ‘UnboundLocalError’ using code examples.

Understanding UnboundLocalError

UnboundLocalError is a common error in Python that occurs when a local variable is referenced before it has been assigned a value. This error can be confusing for beginners because it is not always clear why it occurs or how to fix it. In this section, we will explore what UnboundLocalError is, why it occurs, and how to identify it.

What is UnboundLocalError?

UnboundLocalError is an exception that occurs when a local variable is referenced before it has been assigned a value. In Python, variables can have either a local or global scope. Local variables are defined within a function and are only accessible within that function. Global variables, on the other hand, are defined outside of a function and can be accessed by any function within the module.

Why Does UnboundLocalError Occur?

UnboundLocalError occurs when a local variable is referenced before it has been assigned a value. This can happen if the variable is defined within a function but is not assigned a value before it is referenced. It can also happen if the variable is defined as a global variable but is not explicitly declared as such using the global statement.

How to Identify UnboundLocalError

UnboundLocalError can be identified by the traceback message that is generated when the error occurs. The traceback message will indicate the line number where the error occurred and provide information about the variable that caused the error.

To fix UnboundLocalError, you need to ensure that all local variables are assigned a value before they are referenced. You can also use the global statement to explicitly declare a variable as a global variable, allowing it to be accessed by any function within the module.

In conclusion, UnboundLocalError is a common error in Python that occurs when a local variable is referenced before it has been assigned a value. To fix this error, you need to ensure that all local variables are assigned a value before they are referenced and use the global statement to declare global variables. By understanding UnboundLocalError and how to fix it, you can write more robust and error-free Python code.

Fixing UnboundLocalError

If you are a Python developer, you may have encountered the UnboundLocalError error while working with local variables or functions. This error occurs when a local variable is referenced before it is assigned a value within a function. In this section, we will discuss how to fix UnboundLocalError in Python.

Solutions for UnboundLocalError

There are several ways to fix UnboundLocalError in Python. One solution is to explicitly declare the variable as global using the global keyword. This will make the variable a global variable instead of a local variable. Here is an example:

In this example, we declared num as a global variable inside the test() function using the global keyword. This allowed us to access and modify the value of num inside the function without raising an UnboundLocalError .

Another solution is to use the int() function to initialize the variable with a value of 0. This will ensure that the variable has a value before it is referenced. Here is an example:

In this example, we used the int() function to initialize num with a value of 0. This prevented the UnboundLocalError from being raised when we referenced num before assigning it a value.

How to Avoid UnboundLocalError

To avoid UnboundLocalError , it is important to understand the concept of local scope and local names in Python. Local scope refers to the area of a program where a variable is defined and can be accessed. Local names refer to the variables defined within a function.

To prevent UnboundLocalError , you should always make sure to assign a value to a local variable before referencing it within a function. You should also avoid using the same name for both global and local variables, as this can cause confusion and lead to errors.

Another way to avoid UnboundLocalError is to use lexical scoping. This means defining a function within another function, which allows the inner function to access the variables of the outer function. This can help prevent UnboundLocalError by ensuring that all variables are defined and assigned a value before they are referenced.

In conclusion, UnboundLocalError is a common error in Python that can be fixed by explicitly declaring variables as global or initializing them with a value using the int() function. To avoid UnboundLocalError , it is important to understand the concept of local scope and local names, and to assign values to local variables before referencing them within a function.

In conclusion, understanding the ‘UnboundLocalError’ in Python is essential for any programmer. This error occurs when a local variable is referenced before it has been assigned a value within a function. It can be frustrating to deal with, but fortunately, there are several ways to fix it.

One common solution is to use the global keyword to declare the variable as global within the function. This allows the function to access the variable outside of its scope. Another solution is to use default arguments in the function definition to initialize the variable with a default value.

It is important to note that this error is a runtime error and can only be detected when the code is executed. Therefore, it is crucial to test your code thoroughly to catch any ‘UnboundLocalError’ before deploying it.

Python is a versatile programming language that is widely used in various fields. Understanding the ‘UnboundLocalError’ and how to fix it is a crucial aspect of programming in Python. By following the tips and tricks outlined in this article, you can avoid this error and write efficient and effective code.

In summary, this guide has covered the basics of the ‘UnboundLocalError’ in Python, including its causes and solutions. We have seen how to use the global keyword and default arguments to fix this error. Hopefully, this article has been helpful in your programming journey, and you can now write better code in Python.

Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

If you try to reference a local variable before assigning a value to it within the body of a function, you will encounter the UnboundLocalError: local variable referenced before assignment.

The preferable way to solve this error is to pass parameters to your function, for example:

Alternatively, you can declare the variable as global to access it while inside a function. For example,

This tutorial will go through the error in detail and how to solve it with code examples .

Table of contents

What is scope in python, unboundlocalerror: local variable referenced before assignment, solution #1: passing parameters to the function, solution #2: use global keyword, solution #1: include else statement, solution #2: use global keyword.

Scope refers to a variable being only available inside the region where it was created. A variable created inside a function belongs to the local scope of that function, and we can only use that variable inside that function.

A variable created in the main body of the Python code is a global variable and belongs to the global scope. Global variables are available within any scope, global and local.

UnboundLocalError occurs when we try to modify a variable defined as local before creating it. If we only need to read a variable within a function, we can do so without using the global keyword. Consider the following example that demonstrates a variable var created with global scope and accessed from test_func :

If we try to assign a value to var within test_func , the Python interpreter will raise the UnboundLocalError:

This error occurs because when we make an assignment to a variable in a scope, that variable becomes local to that scope and overrides any variable with the same name in the global or outer scope.

var +=1 is similar to var = var + 1 , therefore the Python interpreter should first read var , perform the addition and assign the value back to var .

var is a variable local to test_func , so the variable is read or referenced before we have assigned it. As a result, the Python interpreter raises the UnboundLocalError.

Example #1: Accessing a Local Variable

Let’s look at an example where we define a global variable number. We will use the increment_func to increase the numerical value of number by 1.

Let’s run the code to see what happens:

The error occurs because we tried to read a local variable before assigning a value to it.

We can solve this error by passing a parameter to increment_func . This solution is the preferred approach. Typically Python developers avoid declaring global variables unless they are necessary. Let’s look at the revised code:

We have assigned a value to number and passed it to the increment_func , which will resolve the UnboundLocalError. Let’s run the code to see the result:

We successfully printed the value to the console.

We also can solve this error by using the global keyword. The global statement tells the Python interpreter that inside increment_func , the variable number is a global variable even if we assign to it in increment_func . Let’s look at the revised code:

Let’s run the code to see the result:

Example #2: Function with if-elif statements

Let’s look at an example where we collect a score from a player of a game to rank their level of expertise. The variable we will use is called score and the calculate_level function takes in score as a parameter and returns a string containing the player’s level .

In the above code, we have a series of if-elif statements for assigning a string to the level variable. Let’s run the code to see what happens:

The error occurs because we input a score equal to 40 . The conditional statements in the function do not account for a value below 55 , therefore when we call the calculate_level function, Python will attempt to return level without any value assigned to it.

We can solve this error by completing the set of conditions with an else statement. The else statement will provide an assignment to level for all scores lower than 55 . Let’s look at the revised code:

In the above code, all scores below 55 are given the beginner level. Let’s run the code to see what happens:

We can also create a global variable level and then use the global keyword inside calculate_level . Using the global keyword will ensure that the variable is available in the local scope of the calculate_level function. Let’s look at the revised code.

In the above code, we put the global statement inside the function and at the beginning. Note that the “default” value of level is beginner and we do not include the else statement in the function. Let’s run the code to see the result:

Congratulations on reading to the end of this tutorial! The UnboundLocalError: local variable referenced before assignment occurs when you try to reference a local variable before assigning a value to it. Preferably, you can solve this error by passing parameters to your function. Alternatively, you can use the global keyword.

If you have if-elif statements in your code where you assign a value to a local variable and do not account for all outcomes, you may encounter this error. In which case, you must include an else statement to account for the missing outcome.

For further reading on Python code blocks and structure, go to the article: How to Solve Python IndentationError: unindent does not match any outer indentation level .

Go to the  online courses page on Python  to learn more about Python for data science and machine learning.

Have fun and happy researching!

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Python 3: UnboundLocalError: local variable referenced before assignment

This error occurs when you are trying to access a variable before it has been assigned a value. Here is an example of a code snippet that would raise this error:

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The error message will be:

In this example, the variable x is being accessed before it is assigned a value, which is causing the error. To fix this, you can either move the assignment of the variable x before the print statement, or give it an initial value before the print statement.

Both will work without any error.

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Python local variable referenced before assignment Solution

When you start introducing functions into your code, you’re bound to encounter an UnboundLocalError at some point. This error is raised when you try to use a variable before it has been assigned in the local context .

In this guide, we talk about what this error means and why it is raised. We walk through an example of this error in action to help you understand how you can solve it.

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What is unboundlocalerror: local variable referenced before assignment.

Trying to assign a value to a variable that does not have local scope can result in this error:

Python has a simple rule to determine the scope of a variable. If a variable is assigned in a function , that variable is local. This is because it is assumed that when you define a variable inside a function you only need to access it inside that function.

There are two variable scopes in Python: local and global. Global variables are accessible throughout an entire program; local variables are only accessible within the function in which they are originally defined.

Let’s take a look at how to solve this error.

An Example Scenario

We’re going to write a program that calculates the grade a student has earned in class.

We start by declaring two variables:

These variables store the numerical and letter grades a student has earned, respectively. By default, the value of “letter” is “F”. Next, we write a function that calculates a student’s letter grade based on their numerical grade using an “if” statement :

Finally, we call our function:

This line of code prints out the value returned by the calculate_grade() function to the console. We pass through one parameter into our function: numerical. This is the numerical value of the grade a student has earned.

Let’s run our code and see what happens:

An error has been raised.

The Solution

Our code returns an error because we reference “letter” before we assign it.

We have set the value of “numerical” to 42. Our if statement does not set a value for any grade over 50. This means that when we call our calculate_grade() function, our return statement does not know the value to which we are referring.

We do define “letter” at the start of our program. However, we define it in the global context. Python treats “return letter” as trying to return a local variable called “letter”, not a global variable.

We solve this problem in two ways. First, we can add an else statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

If you are using an “if” statement where you declare a variable, you should make sure there is an “else” statement in place. This will make sure that even if none of your if statements evaluate to True, you can still set a value for the variable with which you are going to work.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our calculate_grade() function. However, this approach is likely to lead to more confusing code and other issues. In general, variables should not be declared using “global” unless absolutely necessary . Your first, and main, port of call should always be to make sure that a variable is correctly defined.

In the example above, for instance, we did not check that the variable “letter” was defined in all use cases.

That’s it! We have fixed the local variable error in our code.

The UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. You can solve this error by ensuring that a local variable is declared before you assign it a value.

Now you’re ready to solve UnboundLocalError Python errors like a professional developer !

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How to Fix Local Variable Referenced Before Assignment Error in Python

Table of Contents

Fixing local variable referenced before assignment error.

In Python , when you try to reference a variable that hasn't yet been given a value (assigned), it will throw an error.

That error will look like this:

In this post, we'll see examples of what causes this and how to fix it.

Let's begin by looking at an example of this error:

If you run this code, you'll get

The issue is that in this line:

We are defining a local variable called value and then trying to use it before it has been assigned a value, instead of using the variable that we defined in the first line.

If we want to refer the variable that was defined in the first line, we can make use of the global keyword.

The global keyword is used to refer to a variable that is defined outside of a function.

Let's look at how using global can fix our issue here:

Global variables have global scope, so you can referenced them anywhere in your code, thus avoiding the error.

If you run this code, you'll get this output:

In this post, we learned at how to avoid the local variable referenced before assignment error in Python.

The error stems from trying to refer to a variable without an assigned value, so either make use of a global variable using the global keyword, or assign the variable a value before using it.

Thanks for reading!

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How to Solve Error - Local Variable Referenced Before Assignment in Python

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Check the Variable Scope to Fix the local variable referenced before assignment Error in Python

Initialize the variable before use to fix the local variable referenced before assignment error in python, use conditional assignment to fix the local variable referenced before assignment error in python.

This article delves into various strategies to resolve the common local variable referenced before assignment error. By exploring methods such as checking variable scope, initializing variables before use, conditional assignments, and more, we aim to equip both novice and seasoned programmers with practical solutions.

Each method is dissected with examples, demonstrating how subtle changes in code can prevent this frequent error, enhancing the robustness and readability of your Python projects.

The local variable referenced before assignment occurs when some variable is referenced before assignment within a function’s body. The error usually occurs when the code is trying to access the global variable.

The primary purpose of managing variable scope is to ensure that variables are accessible where they are needed while maintaining code modularity and preventing unexpected modifications to global variables.

We can declare the variable as global using the global keyword in Python. Once the variable is declared global, the program can access the variable within a function, and no error will occur.

The below example code demonstrates the code scenario where the program will end up with the local variable referenced before assignment error.

In this example, my_var is a global variable. Inside update_var , we attempt to modify it without declaring its scope, leading to the Local Variable Referenced Before Assignment error.

We need to declare the my_var variable as global using the global keyword to resolve this error. The below example code demonstrates how the error can be resolved using the global keyword in the above code scenario.

In the corrected code, we use the global keyword to inform Python that my_var references the global variable.

When we first print my_var , it displays the original value from the global scope.

After assigning a new value to my_var , it updates the global variable, not a local one. This way, we effectively tell Python the scope of our variable, thus avoiding any conflicts between local and global variables with the same name.

Ensure that the variable is initialized with some value before using it. This can be done by assigning a default value to the variable at the beginning of the function or code block.

The main purpose of initializing variables before use is to ensure that they have a defined state before any operations are performed on them. This practice is not only crucial for avoiding the aforementioned error but also promotes writing clear and predictable code, which is essential in both simple scripts and complex applications.

In this example, the variable total is used in the function calculate_total without prior initialization, leading to the Local Variable Referenced Before Assignment error. The below example code demonstrates how the error can be resolved in the above code scenario.

In our corrected code, we initialize the variable total with 0 before using it in the loop. This ensures that when we start adding item values to total , it already has a defined state (in this case, 0).

This initialization is crucial because it provides a starting point for accumulation within the loop. Without this step, Python does not know the initial state of total , leading to the error.

Conditional assignment allows variables to be assigned values based on certain conditions or logical expressions. This method is particularly useful when a variable’s value depends on certain prerequisites or states, ensuring that a variable is always initialized before it’s used, thereby avoiding the common error.

In this example, message is only assigned within the if and elif blocks. If neither condition is met (as with guest ), the variable message remains uninitialized, leading to the Local Variable Referenced Before Assignment error when trying to print it.

The below example code demonstrates how the error can be resolved in the above code scenario.

In the revised code, we’ve included an else statement as part of our conditional logic. This guarantees that no matter what value user_type holds, the variable message will be assigned some value before it is used in the print function.

This conditional assignment ensures that the message is always initialized, thereby eliminating the possibility of encountering the Local Variable Referenced Before Assignment error.

Throughout this article, we have explored multiple approaches to address the Local Variable Referenced Before Assignment error in Python. From the nuances of variable scope to the effectiveness of initializations and conditional assignments, these strategies are instrumental in developing error-free code.

The key takeaway is the importance of understanding variable scope and initialization in Python. By applying these methods appropriately, programmers can not only resolve this specific error but also enhance the overall quality and maintainability of their code, making their programming journey smoother and more rewarding.

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4 ways to fix local variable referenced before assignment error in python, resolving the local variable referenced before assignment error in python.

Python is one of the world’s most popular programming languages due to its simplicity, readability, and versatility. Despite its many advantages, when coding in Python, one may encounter various errors, with the most common being the “local variable referenced before assignment” error.

Even the most experienced Python developers have encountered this error at some point in their programming career. In this article, we will look at four effective strategies for resolving the local variable referenced before assignment error in Python.

Strategy 1: Assigning a Value before Referencing

The first strategy is to assign a value to a variable before referencing it. The error occurs when the variable is referenced before it is assigned a value.

This problem can be avoided by initializing the variable before referencing it. For example, let us consider the snippet below:

In the snippet above, the variables x and y are not assigned values before they are referenced in the print statement. Therefore, we will get a local variable “referenced before assignment” error.

To resolve this error, we must initialize the variables before referencing them. We can avoid this error by assigning a value to x and y before they are referenced, as shown below:

Strategy 2: Using the Global Keyword

In Python, variables declared inside a function are considered local variables. Thus, they are separate from other variables declared outside of the function.

If we want to use a variable outside of the function, we must use the global keyword. Using the global keyword tells Python that you want to use the variable that was defined globally, not locally.

For example:

In the code snippet above, the global keyword tells Python to use the variable x defined outside of the function rather than a local variable named x . Thus, Python will output 30.

Strategy 3: Adding Input Parameters for Functions

Another way to avoid the local variable referenced before assignment error is by adding input parameters to functions.

In the code snippet above, x and y are variables that are passed into the add_numbers function as arguments.

This approach allows us to avoid the local variable referenced before assignment error because the variables are being passed into the function as input parameters.

Strategy 4: Initializing Variables before Loops or Conditionals

Finally, it’s also a good practice to initialize the variables before loops or conditionals.

If you are defining a variable within a loop, you must initialize it before the loop starts. This way, the variable already exists, and we can update the value inside the loop.

In the code snippet above, the variable sum has been initialized with the value of 0 before the loop runs. Thus, we can update and use the variable inside the loop.

In conclusion, the “local variable referenced before assignment” error is a common issue in Python. However, with the strategies discussed in this article, you can avoid the error and write clean Python code.

Remember to initialize your variables, use the global keyword, add input parameters in functions, and initialize variables before loops or conditionals. By following these techniques, your Python code will be error-free and much easier to manage.

In essence, this article has provided four key strategies for resolving the “local variable referenced before assignment” error that is common in Python. These strategies include initializing variables before referencing, using the global keyword, adding input parameters to functions, and initializing variables before loops or conditionals.

These techniques help to ensure clean code that is free from errors. By implementing these strategies, developers can improve their code quality and avoid time-wasting errors that can occur in their work.

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What is UnboundLocalError: local variable referenced before assignment?

Trying to assign a value to a variable that does not have local scope can result in this error:

Python has a simple rule to determine the scope of a variable. To clarify, a variable is assigned in a function, that variable is local. Because it is assumed that when you define a variable inside a function, you only need to access it inside that function.

There are two variable scopes in Python: local and global. Global variables are accessible throughout an entire program. Whereas, local variables are only accessible within the function in which they are originally defined.

An example of Local variable referenced before assignment

We’re going to write a program that calculates the grade a student has earned in class.

Firstly, we start by declaring two variables:

These variables store the numerical and letter grades a student has earned, respectively. By default, the value of “letter” is “F”. Then, we write a function that calculates a student’s letter grade based on their numerical grade using an “if” statement:

Finally, we call our function:

This line of code prints out the value returned by the  calculate_grade()  function to the console. We pass through one parameter into our function: numerical. This is the numerical value of the grade a student has earned.

Let’s run our code of Local variable referenced before assignment and see what happens:

Here is an error!

The Solution of Local variable referenced before assignment

The code returns an error: Unboundlocalerror local variable referenced before assignment because we reference “letter” before we assign it.

We have set the value of “numerical” to 42. Our  if  statement does not set a value for any grade over 50. This means that when we call our  calculate_grade()  function, our return statement does not know the value to which we are referring.

Moreover, we do define “letter” at the start of our program. However, we define it in the global context. Because Python treats “return letter” as trying to return a local variable called “letter”, not a global variable.

Therefore, this problem of variable referenced before assignment could be solved in two ways. Firstly, we can add an  else  statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade. This approach is good because it lets us keep “letter” in the local context. To clarify, we could even remove the “letter = “F”” statement from the top of our code because we do not use it in the global context.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our  calculate_grade()  function:

We use the “global” keyword at the start of our function.

This keyword changes the scope of our variable to a global variable. This means the “return” statement will no longer treat “letter” like a local variable. Let’s run our code. Our code returns: F.

The code works successfully! Let’s try it using a different grade number by setting the value of “numerical” to a new number:

Our code returns: B.

Finally, we have fixed the local variable referenced before assignment error in the code.

To sum up, as you can see, the UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. Then, you can solve this error by ensuring that a local variable is declared before you assign it a value. Moreover, if a variable is declared globally that you want to access in a function, you can use the “global” keyword to change its value. In case you have any inquiry, let’s CONTACT US . With a lot of experience in Mobile app development services , we will surely solve it for you instantly.

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UnboundLocalError: local variable 'x' referenced before assignment - Python Debugging | Summary and Q&A

Unbound Local Error occurs when there is a variable reference before its assignment, but it can be resolved by declaring the variable as global.

Install to Summarize YouTube Videos and Get Transcripts

Key insights.

• ❓ Unbound Local Error is similar to Name Error and both involve variable references before assignment.
• 🌐 Declaring a variable as global can resolve Unbound Local Error.
• 😕 Unbound Local Error can be confusing for beginners learning programming.
• 🌐 The "global" keyword is used to specify a variable as global within a function.
• ❓ Unbound Local Error can prevent the modification of variables within functions.
• ❓ Understanding Unbound Local Error is essential for efficient debugging in Python.
• 🌐 Global variables can be accessed and modified within functions by declaring them as global.

hello and welcome to the second Python debugging tutorial series in the last video we were talking about name error and in this tutorial we're going to be talking about unbound local error and they're kind of similar in what causes this error and both of them basically have a variable reference before the assignment so let's go ahead and run throug... Read More

Questions & Answers

Q: what causes unbound local error in python.

Unbound Local Error occurs when a variable is referenced before it is assigned a value within a function.

Q: How can Unbound Local Error be fixed?

To fix Unbound Local Error, the variable causing the error can be declared as global using the "global" keyword before its assignment within the function.

Q: How does declaring a variable as global help in resolving Unbound Local Error?

Declaring a variable as global allows the function to access the global version of the variable, preventing the error caused by referencing it before assignment within the function.

Q: Can Unbound Local Error occur outside of a function?

No, Unbound Local Error specifically occurs within a function when a variable is referenced before it is assigned a value within that function.

Summary & Key Takeaways

This tutorial explains the concept of Unbound Local Error in Python, which is similar to Name Error.

Unbound Local Error occurs when a variable is referenced before it is assigned a value.

To fix Unbound Local Error, the variable can be declared as global using the "global" keyword.

Share This Summary 📚

Summarize youtube videos and get video transcripts with 1-click, explore more summaries from sentdex 📚.

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How To Fix Unbound Local Errors In Python

Rahul Kumar Yadav

The beginner users of python may get confused when they see an Unbound Local Error problem in Python. To save them from this headache we have provided the cause, and the fixes of this error, which often occurs while coding in Python.

How UnboundLocalError: local variable referenced before assignment occurs

UnboundLocalError: local variable referenced before assignment , error occurs when we are trying to reference the variable inside a function before assigning any value to it.

This error occurs only when the local variable which is in focus is being assigned some value in the function, and we have referenced it before the assignment operation.

Let us take the help of the code to see what I am trying to say:-

The above example will show an error in line 3, where it says UnboundLocalError: local_variable 'string' referenced before assignment .

The above error occurs because there is both assignment and reference of the variable string, and thus, program creates a local variable named as string for the function example, but it is referenced before any value has been assigned(in the line 3, as we are trying to use the value of string to update the value of string) and since, till then there is no value for local variable string in the function, it shows unbound local error.

Had the code been, as given below:-

It would have simply displayed the output:-

As in this case, there is no assignment of the value to the string variable, and thus, no local variable string is created, and thus, it uses the value in the global scope to display.

The problem of UnboundLocalError occurs only when there is an assignment operation of the variable having the same name as outside the function and inside the function, and the variable is referenced before the assignment operation has taken place.

How to Fix UnboundLocalError error

Only way to avoid this error is to assign the variable before referencing it, or to modify it inside the function with a different name and then return that modified value and assign it to the variable outside the function.

Various Ways of Fixing the Above Error

Sometimes we need to use the value of the variable outside the function, and thus it may require referencing before assignment.

There are some ways by which we can bypass this problem, where we need to reference before assignment.

• Using global keyword:-

The output of the above code is:-

Using the global keyword we can easily circumvent the problem of no referencing before assignment, as then it would be able to use the global variable for further operations inside the function.

• We can pass the value of variable outside the function as a parameter to the function:-

Upon passing the value, we will be able to use the variable’s value without declaring it as a global variable, as sometimes it is beneficial to not expose the value to every function.

• Sometimes we want to modify the value of global variable without declaring it as a global variable, we can simply use some other variable, and modify it accordingly and then return its value and assign it to variable whose value we wanted to modify, like in the given example:-

The output of the above code will be :-

In the above code, we have modified the variable string without declaring it as a global variable, by returning the modified value from the function.

• If we are using a nested function and we want to assign a value to the local variables from the outer function, then we can use the “nonlocal” keyword to access that variable inside the nested function.

The output of the above function will be:-

We can see in the output that we have modified the string present in the outer_example() by accessing it inside the inner_example() .

We can also use the methods numbered 2 and 3 along with this one to modify the variable present in the outer function from the inner function.

We just saw the reasons behind the occurrence of Unbound Local Error in Python, along with various fixes which we can use in our function to bypass the problem along with getting our task accomplished.

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Unbound error, local variable referenced before assignment

I'm making a small text game for fun. I want to use a function which is located in a function file that I made called functionala.

The function in question, attack() , does not work and the program crashes with the error:

The new and improved version of the functionala file is what seems to be causing the problem:

• Is the indentation as shown in this question the same as in your code? Your first snippet has excessive indentation for the first line and too little for one print statement. Your second snippet has excess for all lines. Indentation is important in Python! –  Rory Daulton Commented Apr 5, 2017 at 18:52
• 4 So you have two files, and each file imports the other file? While this is not strictly forbidden by Python, it can often lead to confusing error messages or surprising behavior, which seems to be the case here. Try to redesign your code so no circular imports occur. –  Kevin Commented Apr 5, 2017 at 18:52
• @Rory Daulton no, my indentation is wrong in this and it isn't the same in the files. sorry, couldn't find a way to import the code in its original format –  Dominic Morgan Commented Apr 5, 2017 at 18:54
• Possible duplicate of Python circular importing? –  Open AI - Opting Out Commented Apr 5, 2017 at 18:57
• @Kevin alright, I have redesigned my code by removing the import in "functionala". unfortunately, it now says that my variablemonsterhealth is referenced before assignment, even though the variable is assigned before the actual function. –  Dominic Morgan Commented Apr 5, 2017 at 19:02

This seems a cleaner way of doing it, all in a single file. you may want to look at using classes.

From console, call game() to start the game, that's it. The game will end when either monster or you have health <= 0.

NB: Even doing this in a single file, you need to scope the variables to the "main" procedure ( game ), and pass them to the attack function. Otherwise, referring to these names will raise the same error, and you can reproduce this like so:

HOWEVER, and this may be confusing, the following will not raise an error:

Nor will this:

But both of those seem kind of hack-y and it's usually better to pass values from the main procedure to called functions/methods/etc and return appropriate values to the caller.

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