Local variable referenced before assignment in Python
Last updated: Feb 17, 2023 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.
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
Copyright © 2023 Borislav Hadzhiev
[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:
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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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.
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
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.
Fixing local variable referenced before assignment error
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.
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UnboundLocalError: local variable referenced before assignment
This is a question I had when I was doing Jupyter Notebook practice. First, I defined the function get_y() and tested it:
Then I wrote another function:
When I run the testing codes:
The error shows:
UnboundLocalError: local variable ‘get_y’ referenced before assignment
I tried to fix it by replacing “get_y” variable to “y”, and it worked. I googled this error but I am still a little confused: I didn’t find my “get_y” variable referenced before. Is it because there is a conflict between variable name and the function name?
Where is it referenced? Where is it assigned? Use the trace back to help discover one of the endpoints, then trace back to the other.
An UnboundLocalError is raised when a local variable is referenced before it has been assigned. In most cases this will occur when trying to modify a local variable before it is actually assigned within the local scope. Python doesn’t have variable declarations, so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function, that variable is considered local.
Python has lexical scoping by default, which means that although an enclosed scope can access values in its enclosing scope, it cannot modify them (unless they’re declared global with the global keyword). A closure binds values in the enclosing environment to names in the local environment. The local environment can then use the bound value, and even reassign that name to something else, but it can’t modify the binding in the enclosing environment. UnboundLocalError happend because when python sees an assignment inside a function then it considers that variable as local variable and will not fetch its value from enclosing or global scope when we execute the function. However, to modify a global variable inside a function, you must use the global keyword.
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UnboundLocalError: local variable 'pool' referenced before assignment #699
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Brandoncbryant commented feb 13, 2022 • edited, edwinnglabs commented feb 13, 2022.
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Local variable referenced before assignment in Python
The “local variable referenced before assignment” error occurs in Python when you try to use a local variable before it has been assigned a value.
This error typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.
Here’s an example to illustrate this error:
In this example, you would encounter the “local variable ‘x’ referenced before assignment” error because you’re trying to print the value of x before it has been assigned a value. To fix this, you should assign a value to x before attempting to access it:
In the corrected version, the local variable x is assigned a value before it’s used, preventing the error.
Keep in mind that Python treats variables inside functions as local unless explicitly stated otherwise using the global keyword (for global variables) or the nonlocal keyword (for variables in nested functions).
If you encounter this error and you’re sure that the variable should have been assigned a value before its use, double-check your code for any logical errors or typos that might be causing the variable to not be assigned properly.
Using the global keyword
If you have a global variable named letter and you try to modify it inside a function without declaring it as global, you will get error.
This is because Python assumes that any variable that is assigned a value inside a function is a local variable, unless you explicitly tell it otherwise.
To fix this error, you can use the global keyword to indicate that you want to use the global variable:
Using nonlocal keyword
The nonlocal keyword is used to work with variables inside nested functions, where the variable should not belong to the inner function. It allows you to modify the value of a non-local variable in the outer scope.
For example, if you have a function outer that defines a variable x , and another function inner inside outer that tries to change the value of x , you need to use the nonlocal keyword to tell Python that you are referring to the x defined in outer , not a new local variable in inner .
Here is an example of how to use the nonlocal keyword:
If you don’t use the nonlocal keyword, Python will create a new local variable x in inner , and the value of x in outer will not be changed:
Sometimes you may suddenly start getting UnboundLocalError in your code which was working perfectly just minutes ago. And what you did is just added an assignment statement.
Lets say your code is as below, which is working fine.
Now you added an assignment statement inside printx function where you are increasing the value of global variable x . You will start getting UnboundLocalError .
What caused UnboundLocalError?
Lets understand few things first.
In python all the variables inside a function are global if they are not assigned any value to them. That means if a variable is only referenced inside a function, it is global. However if we assign any value to a variable inside a function, its scope becomes local to that unless explicitly declared global.
In first example, we are not assigning any value to x variable inside function and just referencing it to print. Hence x is global.
In second example, we assigned the value 7 to x, hence x's scope is local to function and we tried to print it before assigning any value to x.
Similarly, you will get UnboundLocalError in scenario similar to below example.
UnboundLocalError can be solved by changing the scope of the variable which is complaining. You need to explicitly declare the variable global.
Variable x's scope in function printx is global. You can verify the same by printing the value of x in terminal and it will be 6.
Remember that same thing can be done using nonlocal keyword. But mind it that nonlocal binds the variable to nearest enclosing scope except global. Lets understand this with an example.
Output will be:
So take away is:
1. Use global keyword with a variable if only you are assigning value to that variable inside a function and you want to overwrite the value of variable declared in global scope.
2. Use nonlocal keyword in nested scopes.
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