Fix "local variable referenced before assignment" in Python

what does local variable referenced before assignment mean

Introduction

If you're a Python developer, you've probably come across a variety of errors, like the "local variable referenced before assignment" error. This error can be a bit puzzling, especially for beginners and when it involves local/global variables.

Today, we'll explain this error, understand why it occurs, and see how you can fix it.

The "local variable referenced before assignment" Error

The "local variable referenced before assignment" error in Python is a common error that occurs when a local variable is referenced before it has been assigned a value. This error is a type of UnboundLocalError , which is raised when a local variable is referenced before it has been assigned in the local scope.

Here's a simple example:

Running this code will throw the "local variable 'x' referenced before assignment" error. This is because the variable x is referenced in the print(x) statement before it is assigned a value in the local scope of the foo function.

Even more confusing is when it involves global variables. For example, the following code also produces the error:

But wait, why does this also produce the error? Isn't x assigned before it's used in the say_hello function? The problem here is that x is a global variable when assigned "Hello ". However, in the say_hello function, it's a different local variable, which has not yet been assigned.

We'll see later in this Byte how you can fix these cases as well.

Fixing the Error: Initialization

One way to fix this error is to initialize the variable before using it. This ensures that the variable exists in the local scope before it is referenced.

Let's correct the error from our first example:

In this revised code, we initialize x with a value of 1 before printing it. Now, when you run the function, it will print 1 without any errors.

Fixing the Error: Global Keyword

Another way to fix this error, depending on your specific scenario, is by using the global keyword. This is especially useful when you want to use a global variable inside a function.

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Here's how:

In this snippet, we declare x as a global variable inside the function foo . This tells Python to look for x in the global scope, not the local one . Now, when you run the function, it will increment the global x by 1 and print 1 .

Similar Error: NameError

An error that's similar to the "local variable referenced before assignment" error is the NameError . This is raised when you try to use a variable or a function name that has not been defined yet.

Running this code will result in a NameError :

In this case, we're trying to print the value of y , but y has not been defined anywhere in the code. Hence, Python raises a NameError . This is similar in that we are trying to use an uninitialized/undefined variable, but the main difference is that we didn't try to initialize y anywhere else in our code.

Variable Scope in Python

Understanding the concept of variable scope can help avoid many common errors in Python, including the main error of interest in this Byte. But what exactly is variable scope?

In Python, variables have two types of scope - global and local. A variable declared inside a function is known as a local variable, while a variable declared outside a function is a global variable.

Consider this example:

In this code, x is a global variable, and y is a local variable. x can be accessed anywhere in the code, but y can only be accessed within my_function . Confusion surrounding this is one of the most common causes for the "variable referenced before assignment" error.

In this Byte, we've taken a look at the "local variable referenced before assignment" error and another similar error, NameError . We also delved into the concept of variable scope in Python, which is an important concept to understand to avoid these errors. If you're seeing one of these errors, check the scope of your variables and make sure they're being assigned before they're being used.

what does local variable referenced before assignment mean

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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

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!

what does local variable referenced before assignment mean

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Local variable referenced before assignment in Python

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

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# 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 .

unboundlocalerror local variable name referenced before assignment

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.

mark variable as global

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 global variables in functions

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.

variables declared in function cannot be accessed in 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.

return value from the function

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.

pass global variable as argument to 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.

assign value to local variable from outer scope

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.

not using nonlocal prints 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

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

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

what does local variable referenced before assignment mean

One error you might encounter when running Python code is:

This error commonly occurs when you reference a variable inside a function without first assigning it a value.

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

Let me show you an example that causes this error and how I fix it in practice.

How to reproduce this error

Suppose you have a variable called name declared in your Python code as follows:

Next, you created a function that uses the name variable as shown below:

When you execute the code above, you’ll get this error:

This error occurs because you both assign and reference a variable called name inside the function.

Python thinks you’re trying to assign the local variable name to name , which is not the case here because the original name variable we declared is a global variable.

How to fix this error

To resolve this error, you can change the variable’s name inside the function to something else. For example, name_with_title should work:

As an alternative, you can specify a name parameter in the greet() function to indicate that you require a variable to be passed to the function.

When calling the function, you need to pass a variable as follows:

This code allows Python to know that you intend to use the name variable which is passed as an argument to the function as part of the newly declared name variable.

Still, I would say that you need to use a different name when declaring a variable inside the function. Using the same name might confuse you in the future.

Here’s the best solution to the error:

Now it’s clear that we’re using the name variable given to the function as part of the value assigned to name_with_title . Way to go!

The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable.

To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function.

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

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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:

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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.

How to Solve Error - Local Variable Referenced Before Assignment 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.

python local variable referenced before assignment - output 1

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.

python local variable referenced before assignment - output 2

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.

python local variable referenced before assignment - output 3

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.

[SOLVED] Local Variable Referenced Before Assignment

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

[Fixed] typeerror can’t compare datetime.datetime to datetime.date

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Local Variable Referenced Before Assignment Error

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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[Fixed] nameerror: name Unicode is not defined

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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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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:

“`python

add_numbers():

print(x + y)

add_numbers()

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.

def add_numbers(x, y):

add_numbers(10, 20)

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.

my_list = [1, 2, 3, 4, 5]

for number in my_list:

sum += number

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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Local Variable ‘i’ Referenced Before Assignment: What It Is and How to Fix It

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Have you ever been working on a coding project and gotten the dreaded error message local variable i referenced before assignment? This error can be a real pain to troubleshoot, especially if youre not sure what it means. In this article, well take a look at what this error means and how to fix it. Well also discuss some best practices for writing code that avoids this error in the first place.

So, what does it mean when you get the error local variable i referenced before assignment? It means that youre trying to use a variable before youve assigned it a value. This is a common mistake, and it can cause your code to behave in unexpected ways.

To fix this error, you need to make sure that you assign a value to the variable before you try to use it. You can do this by using the assignment operator (=), like this:

This will assign the value 0 to the variable i. Now, you can use the variable i without getting an error.

Here are some best practices for writing code that avoids the local variable i referenced before assignment error:

  • Declare all variables at the beginning of the function. This will help you to avoid accidentally using a variable before its been assigned a value.
  • Use the assignment operator (=) to assign values to variables. This will help to ensure that variables are initialized with values before theyre used.
  • Use descriptive variable names. This will help you to avoid accidentally using the wrong variable.

By following these best practices, you can help to avoid the local variable i referenced before assignment error and write more reliable code.

In this tutorial, you will learn about local variables and how to avoid referencing them before they have been assigned a value. You will also learn how to debug errors caused by referencing local variables before assignment.

What is a local variable?

A local variable is a variable that is declared within a function or block of code. Local variables are only accessible within the function or block of code in which they are declared. This means that they cannot be accessed from outside of the function or block of code.

What does it mean for a local variable to be referenced before assignment?

A local variable is referenced before assignment when it is used in an expression before it has been assigned a value. This can cause a compile-time error or a runtime error.

A compile-time error occurs when the compiler detects that the local variable has not been assigned a value. The compiler will then generate an error message and the program will not compile.

A runtime error occurs when the program tries to use the local variable before it has been assigned a value. The program will then crash and you will see an error message in the console.

How to avoid referencing local variables before assignment

There are a few ways to avoid referencing local variables before assignment.

  • Use the `const` keyword. The `const` keyword can be used to declare a local variable as constant. This means that the variable cannot be changed after it has been assigned a value. This can help to prevent errors caused by referencing local variables before assignment.
  • Use the `let` keyword. The `let` keyword can be used to declare a local variable. Unlike the `const` keyword, the `let` keyword allows the variable to be changed after it has been assigned a value. However, the `let` keyword will still generate an error if the variable is referenced before it has been assigned a value.
  • Use the `var` keyword. The `var` keyword can be used to declare a local variable. Unlike the `const` and `let` keywords, the `var` keyword does not generate an error if the variable is referenced before it has been assigned a value. However, it is generally considered bad practice to use the `var` keyword because it can lead to errors.

How to debug errors caused by referencing local variables before assignment

If you receive an error message that indicates that a local variable has been referenced before assignment, you can use the following steps to debug the error:

1. Check the code to see where the local variable is being referenced. 2. Make sure that the local variable has been assigned a value before it is being referenced. 3. If the local variable has not been assigned a value, add an assignment statement to the code before the local variable is referenced.

In this tutorial, you learned about local variables and how to avoid referencing them before they have been assigned a value. You also learned how to debug errors caused by referencing local variables before assignment.

By following these tips, you can help to prevent errors in your code and improve the performance of your programs.

3. How to fix the error of local variable referenced before assignment?

There are two ways to fix the error of local variable referenced before assignment:

1. Assign a value to the local variable before it is used.

This is the simplest and most straightforward way to fix the error. For example, the following code will fix the error:

int i; i = 10; printf(“The value of i is %d\n”, i);

2. Declare the local variable as `const`.

Declaring a local variable as `const` means that the variable cannot be changed after it is initialized. This can also be used to fix the error of local variable referenced before assignment. For example, the following code will fix the error:

const int i = 10; printf(“The value of i is %d\n”, i);

4. Examples of local variable referenced before assignment

The following code will cause a compile-time error:

int i; printf(“The value of i is %d\n”, i);

This code will cause an error because the compiler cannot determine the value of `i` before it is used. To fix this error, we can either assign a value to `i` before it is used, or we can declare `i` as `const`.

The following code will cause a runtime error:

This code will cause a runtime error because the value of `i` is not initialized before it is used. To fix this error, we can either assign a value to `i` before it is used, or we can declare `i` as `const`.

The following code will fix the error:

int i = 10; printf(“The value of i is %d\n”, i);

Local variable referenced before assignment is a common error that can be easily fixed. By following the tips in this article, you can avoid this error and write more efficient and error-free code.

Q: What does it mean when I get a compiler error that says “local variable ‘i’ referenced before assignment”?

A: This error occurs when you try to use a variable before it has been assigned a value. For example, the following code will cause an error:

int i; System.out.println(i);

This is because the variable `i` has not been assigned a value yet, so the compiler doesn’t know what to print. To fix this error, you need to assign a value to the variable before you try to use it. For example:

int i = 0; System.out.println(i);

Q: How can I avoid this error?

A: There are a few ways to avoid this error. One way is to make sure that you always assign a value to a variable before you try to use it. Another way is to use the `final` keyword to declare a variable as constant. This means that the variable’s value cannot be changed after it has been initialized, so you won’t get an error if you try to use it before it has been assigned a value.

Q: What other errors can be caused by referencing a variable before it has been assigned a value?

A: There are a few other errors that can be caused by referencing a variable before it has been assigned a value. One error is a `NullPointerException`. This error occurs when you try to use a variable that has not been initialized, and the variable’s value is `null`. Another error is a `ClassCastException`. This error occurs when you try to cast a variable to a type that it is not compatible with.

Q: How can I debug this error?

A: If you are getting an error that says “local variable ‘i’ referenced before assignment”, you can debug the error by using a debugger. A debugger allows you to step through your code line by line, and it can help you identify the line of code that is causing the error.

Q: What are some best practices for avoiding this error?

A: There are a few best practices that you can follow to avoid this error. First, always assign a value to a variable before you try to use it. Second, use the `final` keyword to declare a variable as constant. Third, use a debugger to help you identify the line of code that is causing the error.

In this blog post, we discussed the error local variable i referenced before assignment. We explained what this error means and how to fix it. We also provided some tips on how to avoid this error in the future.

To summarize, the local variable i referenced before assignment error occurs when you try to use a variable before it has been initialized. To fix this error, you need to initialize the variable before you use it. You can do this by assigning a value to the variable, or by declaring the variable with the const keyword.

  • To avoid this error in the future, you should be careful to initialize your variables before you use them. You can also use the const keyword to declare variables that you do not plan to change.

Here are some key takeaways from this blog post:

  • The local variable i referenced before assignment error occurs when you try to use a variable before it has been initialized.
  • To fix this error, you need to initialize the variable before you use it.
  • You can initialize a variable by assigning a value to it, or by declaring the variable with the const keyword.

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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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what does local variable referenced before assignment mean

Understanding UnboundLocalError in Python

If you're closely following the Python tag on StackOverflow , you'll notice that the same question comes up at least once a week. The question goes on like this:

Why, when run, this results in the following error:

There are a few variations on this question, with the same core hiding underneath. Here's one:

Running the lst.append(5) statement successfully appends 5 to the list. However, substitute it for lst += [5] , and it raises UnboundLocalError , although at first sight it should accomplish the same.

Although this exact question is answered in Python's official FAQ ( right here ), I decided to write this article with the intent of giving a deeper explanation. It will start with a basic FAQ-level answer, which should satisfy one only wanting to know how to "solve the damn problem and move on". Then, I will dive deeper, looking at the formal definition of Python to understand what's going on. Finally, I'll take a look what happens behind the scenes in the implementation of CPython to cause this behavior.

The simple answer

As mentioned above, this problem is covered in the Python FAQ. For completeness, I want to explain it here as well, quoting the FAQ when necessary.

Let's take the first code snippet again:

So where does the exception come from? Quoting the FAQ:

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope.

But x += 1 is similar to x = x + 1 , so it should first read x , perform the addition and then assign back to x . As mentioned in the quote above, Python considers x a variable local to foo , so we have a problem - a variable is read (referenced) before it's been assigned. Python raises the UnboundLocalError exception in this case [1] .

So what do we do about this? The solution is very simple - Python has the global statement just for this purpose:

This prints 11 , without any errors. The global statement tells Python that inside foo , x refers to the global variable x , even if it's assigned in foo .

Actually, there is another variation on the question, for which the answer is a bit different. Consider this code:

This kind of code may come up if you're into closures and other techniques that use Python's lexical scoping rules. The error this generates is the familiar UnboundLocalError . However, applying the "global fix":

Doesn't help - another error is generated: NameError: global name 'x' is not defined . Python is right here - after all, there's no global variable named x , there's only an x in external . It may be not local to internal , but it's not global. So what can you do in this situation? If you're using Python 3, you have the nonlocal keyword. Replacing global by nonlocal in the last snippet makes everything work as expected. nonlocal is a new statement in Python 3, and there is no equivalent in Python 2 [2] .

The formal answer

Assignments in Python are used to bind names to values and to modify attributes or items of mutable objects. I could find two places in the Python (2.x) documentation where it's defined how an assignment to a local variable works.

One is section 6.2 "Assignment statements" in the Simple Statements chapter of the language reference:

Assignment of an object to a single target is recursively defined as follows. If the target is an identifier (name): If the name does not occur in a global statement in the current code block: the name is bound to the object in the current local namespace. Otherwise: the name is bound to the object in the current global namespace.

Another is section 4.1 "Naming and binding" of the Execution model chapter:

If a name is bound in a block, it is a local variable of that block. [...] When a name is used in a code block, it is resolved using the nearest enclosing scope. [...] If the name refers to a local variable that has not been bound, a UnboundLocalError exception is raised.

This is all clear, but still, another small doubt remains. All these rules apply to assignments of the form var = value which clearly bind var to value . But the code snippets we're having a problem with here have the += assignment. Shouldn't that just modify the bound value, without re-binding it?

Well, no. += and its cousins ( -= , *= , etc.) are what Python calls " augmented assignment statements " [ emphasis mine ]:

An augmented assignment evaluates the target (which, unlike normal assignment statements, cannot be an unpacking) and the expression list, performs the binary operation specific to the type of assignment on the two operands, and assigns the result to the original target . The target is only evaluated once. An augmented assignment expression like x += 1 can be rewritten as x = x + 1 to achieve a similar, but not exactly equal effect. In the augmented version, x is only evaluated once. Also, when possible, the actual operation is performed in-place, meaning that rather than creating a new object and assigning that to the target, the old object is modified instead. With the exception of assigning to tuples and multiple targets in a single statement, the assignment done by augmented assignment statements is handled the same way as normal assignments . Similarly, with the exception of the possible in-place behavior, the binary operation performed by augmented assignment is the same as the normal binary operations.

So when earlier I said that x += 1 is similar to x = x + 1 , I wasn't telling all the truth, but it was accurate with respect to binding. Apart for possible optimization, += counts exactly as = when binding is considered. If you think carefully about it, it's unavoidable, because some types Python works with are immutable. Consider strings, for example:

The first line binds x to the value "abc". The second line doesn't modify the value "abc" to be "abcdef". Strings are immutable in Python . Rather, it creates the new value "abcdef" somewhere in memory, and re-binds x to it. This can be seen clearly when examining the object ID for x before and after the += :

Note that some types in Python are mutable. For example, lists can actually be modified in-place:

id(y) didn't change after += , because the object y referenced was just modified. Still, Python re-bound y to the same object [3] .

The "too much information" answer

This section is of interest only to those curious about the implementation internals of Python itself.

One of the stages in the compilation of Python into bytecode is building the symbol table [4] . An important goal of building the symbol table is for Python to be able to mark the scope of variables it encounters - which variables are local to functions, which are global, which are free (lexically bound) and so on.

When the symbol table code sees a variable is assigned in a function, it marks it as local. Note that it doesn't matter if the assignment was done before usage, after usage, or maybe not actually executed due to a condition in code like this:

We can use the symtable module to examine the symbol table information gathered on some Python code during compilation:

This prints:

So we see that x was marked as local in foo . Marking variables as local turns out to be important for optimization in the bytecode, since the compiler can generate a special instruction for it that's very fast to execute. There's an excellent article here explaining this topic in depth; I'll just focus on the outcome.

The compiler_nameop function in Python/compile.c handles variable name references. To generate the correct opcode, it queries the symbol table function PyST_GetScope . For our x , this returns a bitfield with LOCAL in it. Having seen LOCAL , compiler_nameop generates a LOAD_FAST . We can see this in the disassembly of foo :

The first block of instructions shows what x += 1 was compiled to. You will note that already here (before it's actually assigned), LOAD_FAST is used to retrieve the value of x .

This LOAD_FAST is the instruction that will cause the UnboundLocalError exception to be raised at runtime, because it is actually executed before any STORE_FAST is done for x . The gory details are in the bytecode interpreter code in Python/ceval.c :

Ignoring the macro-fu for the moment, what this basically says is that once LOAD_FAST is seen, the value of x is obtained from an indexed array of objects [5] . If no STORE_FAST was done before, this value is still NULL , the if branch is not taken [6] and the exception is raised.

You may wonder why Python waits until runtime to raise this exception, instead of detecting it in the compiler. The reason is this code:

Suppose something_true is a function that returns True , possibly due to some user input. In this case, x = 1 binds x locally, so the reference to it in x += 1 is no longer unbound. This code will then run without exceptions. Of course if something_true actually turns out to return False , the exception will be raised. Python has no way to resolve this at compile time, so the error detection is postponed to runtime.

what does local variable referenced before assignment mean

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COMMENTS

  1. Python 3: UnboundLocalError: local variable referenced before assignment

    File "weird.py", line 5, in main. print f(3) UnboundLocalError: local variable 'f' referenced before assignment. Python sees the f is used as a local variable in [f for f in [1, 2, 3]], and decides that it is also a local variable in f(3). You could add a global f statement: def f(x): return x. def main():

  2. Fix "local variable referenced before assignment" in Python

    Building Your First Convolutional Neural Network With Keras # python # artificial intelligence # machine learning # tensorflow Most resources start with pristine datasets, start at importing and finish at validation.

  3. How to Fix

    Output. Hangup (SIGHUP) Traceback (most recent call last): File "Solution.py", line 7, in <module> example_function() File "Solution.py", line 4, in example_function x += 1 # Trying to modify global variable 'x' without declaring it as global UnboundLocalError: local variable 'x' referenced before assignment Solution for Local variable Referenced Before Assignment in Python

  4. Python local variable referenced before assignment Solution

    Trying to assign a value to a variable that does not have local scope can result in this error: UnboundLocalError: local variable referenced before assignment. 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 ...

  5. How to Fix Local Variable Referenced Before Assignment Error in Python

    value = value + 1 print (value) increment() If you run this code, you'll get. BASH. UnboundLocalError: local variable 'value' referenced before assignment. The issue is that in this line: PYTHON. value = value + 1. 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 ...

  6. Local variable referenced before assignment in Python

    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.

  7. How to fix UnboundLocalError: local variable 'x' referenced before

    The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable. To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function. I hope this tutorial is useful.

  8. Local variable referenced before assignment in Python

    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 ...

  9. Local Variable Referenced Before Assignment in Python

    This tutorial explains the reason and solution of the python error local variable referenced before assignment

  10. [SOLVED] Local Variable Referenced Before Assignment

    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. ... Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable ...

  11. Python UnboundLocalError: local variable referenced before assignment

    UnboundLocalError: local variable referenced before assignment. Example #1: Accessing a Local Variable. Solution #1: Passing Parameters to the Function. Solution #2: Use Global Keyword. Example #2: Function with if-elif statements. Solution #1: Include else statement. Solution #2: Use global keyword. Summary.

  12. 4 Ways to Fix Local Variable Referenced Before Assignment Error in

    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 ...

  13. Local variable referenced before assignment: what it is and how to fix it

    A local variable referenced before assignment is a variable that is used in a program before it has been assigned a value. This can cause errors, as the compiler cannot know what value the variable will have when it is used.

  14. Local variable referenced before assignment?

    @HoKy22: Are you asking why dct[key] = val does not raise a "local variable referenced before assignment" error? The reason is that this is not a bare name assignment. Instead, it causes Python to make the function call dct.__setitem__(key, val). -

  15. Local Variable 'i' Referenced Before Assignment: What It Is and How to

    In this tutorial, you will learn about local variables and how to avoid referencing them before they have been assigned a value. You will also learn how to debug errors caused by referencing local variables before assignment.

  16. Local variable referenced before assignment in Python

    The "Local variable referenced before assignment" appears in Python due to assigning a value to a variable that does not have a local scope. To fix this error, the global keyword, return statement, and nonlocal nested function is used in Python script. The global keywords are used with variables to make it able to access inside and outside ...

  17. UnboundLocalError Local variable Referenced Before Assignment in Python

    Avoid Reassignment of Global Variables. 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.

  18. python : local variable is referenced before assignment

    6. When Python sees that you are assigning to x it forces it to be a local variable name. Now it becomes impossible to see the global x in that function (unless you use the global keyword) So. Case 1) Since there is no local x, you get the global. Case 2) You are assigning to a local x so all references to x in the function will be the local one.

  19. Understanding UnboundLocalError in Python

    I could find two places in the Python (2.x) documentation where it's defined how an assignment to a local variable works. One is section 6.2 "Assignment statements" in the Simple Statements chapter of the language reference: Assignment of an object to a single target is recursively defined as follows. If the target is an identifier (name):

  20. The code says variable referenced before assignment?

    What does 'referenced before assignment' in this case really mean? Because I think that I assign the variable 'lowest' only after 'scores' variable is defined. Any help would be appreciated.

  21. Python Error, Local variable might be referenced before assignment

    Your issue is that result is only defined under if and else-if statements. This means that if every condition in if or else-if statements fails, result will not be defined, yet you will be trying to use its value.

  22. Why am I getting a "local variable 'connector' might be referenced

    UnboundLocalError: local variable 'cur' referenced before assignment after call WinDLL. 1. Variable referenced before assignment in this code. Hot Network Questions Decode Caesar cipher based on a given text Film with a spaceship crew fighting a cyborg who can rebuild themselves. Girl main character