unboundlocalerror local variable 'increment' referenced before assignment

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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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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.

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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

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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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[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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Fix "local variable referenced before assignment" in Python

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.

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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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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 Exception: UnboundLocalError: local variable '…' referenced before assignment

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UnboundLocalError: local variable referenced before assignment

I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment

which isn't clear to me. Any suggestions?

• arcgis-10.1
• unboundlocalerror

• 1 Because if row.getValue("Value") == 1 might be false and so a never gets assigned. –  Nathan W Commented May 20, 2013 at 2:39
• It has value and do gets assigned. I checked it in arcmap interactive python window but can't get it to work in a stand alone script. –  Ibe Commented May 20, 2013 at 2:44
• 1 your loop will also only give you the values of the last loop iteration as you are returning out of the loop and not doing anything with each value. –  Nathan W Commented May 20, 2013 at 2:49
• You could use 3 x elif and an else to see if any values other than 1-4 are encountered. –  PolyGeo ♦ Commented May 20, 2013 at 3:44
• I tried that way as well but still hung up with error. –  Ibe Commented May 20, 2013 at 4:15

This error is pretty much explained here and it helped me to get assignments and return values for all variables.

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UndboundLocalError: local variable referenced before assignment

Hello all, I’m using PsychoPy 2023.2.3 Win 10 x64bits

What I’m trying to do? The experiment will show in the middle of the screen an abstracted stimuli (B1 or B2), and after valid click on it, the stimulus will remain on the middle of the screen and three more stimuli will appear in the cornor of the screen.

I’m having this erro (attached above), a simple error, but I can not see where the error is. Also the experiment isn’t working proberly and is the old version (I don’t know but someone are having troubles with this version of PscyhoPy)? ba_training_block.xlsx (13.8 KB) SMTS.psyexp (91.6 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

You have a routine called sample but you also use that name for your image file in sample_box .

I changed the name of the routine for ‘stimulus_sample’ and manteined the image file in sample_box as ‘sample’. But, the error still remain. But it do not happen all the time, this is very interesting…

Can u give it a look again? (I made some minor changes here)

Here the exp file ba_training_block.xlsx (13.7 KB) SMTS.psyexp (89.7 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

Thanks again

Please could you confirm/show the new error message? Is it definitely still related to sample?

I think you have blank rows in your spreadsheet. The loop claims that there are 19 conditions but I think you only want 12. Without a value for sample_category sample doesn’t get set. With random presentation this will happen at a random point.

Related Topics

【Python】成功解决python报错：UnboundLocalError: local variable ‘xxx‘ referenced before assignment

成功解决python报错：UnboundLocalError: local variable ‘xxx’ referenced before assignment。在Python中， UnboundLocalError 是一种特定的 NameError ，它会在尝试引用一个还未被赋值的局部变量时发生。Python解释器需要知道变量的类型和作用域，因此，在局部作用域内引用一个未被赋值的变量时，就会抛出这个错误。

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【Python】解决Python报错：

1. 什么是unboundlocalerror？, 2. 常见的场景和原因, 方法一：全局变量, 方法二：函数参数, 方法三：局部变量初始化, 方法四：结合条件语句.

在Python编程中， UnboundLocalError: local variable 'xxx' referenced before assignment 是一个常见的错误，尤其是在写函数时可能会遇到。这篇技术博客将详细介绍 UnboundLocalError ，为什么会发生，以及如何解决这个错误。

在Python中， UnboundLocalError 是一种特定的 NameError ，它会在尝试引用一个还未被赋值的局部变量时发生。Python解释器需要知道变量的类型和作用域，因此，在局部作用域内引用一个未被赋值的变量时，就会抛出这个错误。

这是一个简单的代码示例来说明这个错误：

运行以上代码会抛出以下错误：

在这个例子中，Python解释器看到 print(x) 时，寻找局部作用域中的变量 x ，但这个变量在局部作用域内尚未被赋值（虽然在后面有赋值，解释器是从上到下执行代码的）。

理解错误的原因后，可以通过以下几种方式来解决 UnboundLocalError ：

如果变量希望在函数内和函数外都使用，可以将其声明为全局变量：

通过在函数内使用 global 关键字，将 x 声明为全局变量，这样即使在函数内也能访问全局变量 x 。

通过将变量作为参数传递给函数，使得函数内可以访问并使用这个变量：

在这种情况下， x 是函数 my_function 的一个参数，无需在函数内部声明。

在使用变量之前，先初始化该局部变量：

确保在函数内部引用变量之前，该变量已经被赋值。

在复杂的逻辑中，特别是在涉及条件语句时，可以先在函数开始部分初始化变量，确保无论哪条路径都可以正确访问该变量：

在这个例子中，我们确保了变量 x 在函数内部任何地方都能被适当地引用。

• 命名冲突 ：在全局变量和局部变量重名情况下，优先使用局部变量。如果不小心混用，容易引发错误。
• 提前规划变量作用域 ：代码设计时，可以提前规划好变量应该属于哪个作用域，以减少变量冲突和未定义变量的情况。

UnboundLocalError: local variable 'xxx' referenced before assignment 错误是一个常见的初学者错误，但只要理解了Python的变量作用域规则和执行顺序，就可以轻松避开。通过合适的解决方法，如使用全局变量、函数参数、局部变量初始化或结合条件语句，可以高效且清晰地管理变量的使用。

希望这篇文章能帮助你理解和解决这个错误。如果有任何问题或其他建议，欢迎在评论中与我们讨论。Happy coding!

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UnboundLocalError: local variable 'level_num' referenced before assignment #32

liaohu1231 commented May 18, 2024

EMBL-PKU commented May 22, 2024

Sorry, something went wrong.

liaohu1231 commented Jun 2, 2024

hesiyang395 commented Sep 9, 2024

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UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)

When I try this code:

I get an error from the print(c) line that says:

or in some older versions:

If I comment out c += 1 , all the print s are successful.

I don't understand: why does printing a and b work, if c does not? How did c += 1 cause print(c) to fail, even when it comes later in the code?

It seems like the assignment c += 1 creates a local variable c , which takes precedence over the global c . But how can a variable "steal" scope before it exists? Why is c apparently local here?

See also How to use a global variable in a function? for questions that are simply about how to reassign a global variable from within a function, and Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope? for reassigning from an enclosing function (closure).

See Why isn't the 'global' keyword needed to access a global variable? for cases where OP expected an error but didn't get one, from simply accessing a global without the global keyword.

See How can a name be "unbound" in Python? What code can cause an `UnboundLocalError`? for cases where OP expected the variable to be local, but has a logical error that prevents assignment in every case.

• global-variables
• local-variables

14 Answers 14

Python treats variables in functions differently depending on whether you assign values to them from inside or outside the function. If a variable is assigned within a function, it is treated by default as a local variable. Therefore, when you uncomment the line, you are trying to reference the local variable c before any value has been assigned to it.

If you want the variable c to refer to the global c = 3 assigned before the function, put

as the first line of the function.

As for python 3, there is now

that you can use to refer to the nearest enclosing function scope that has a c variable.

• 8 Thanks. Quick question. Does this imply that Python decides the scope of each variable before running a program? Before running a function? –  tba Commented Dec 16, 2008 at 3:46
• 16 The variable scope decision is made by the compiler, which normally runs once when you first start the program. However it is worth keeping in mind that the compiler might also run later if you have "eval" or "exec" statements in your program. –  Greg Hewgill Commented Dec 16, 2008 at 3:48
• 3 Okay thank you. I guess "interpreted language" doesn't imply quite as much as I had thought. –  tba Commented Dec 16, 2008 at 3:53
• 1 Ah that 'nonlocal' keyword was exactly what I was looking for, it seemed Python was missing this. Presumably this 'cascades' through each enclosing scope that imports the variable using this keyword? –  Brendan Commented Nov 17, 2009 at 20:54
• 8 @brainfsck: it is easiest to understand if you make the distinction between "looking up" and "assigning" a variable. Lookup falls back to a higher scope if the name is not found in the current scope. Assignment is always done in the local scope (unless you use global or nonlocal to force global or nonlocal assignment) –  Steven Commented Sep 13, 2011 at 12:00

Python is a little weird in that it keeps everything in a dictionary for the various scopes. The original a,b,c are in the uppermost scope and so in that uppermost dictionary. The function has its own dictionary. When you reach the print(a) and print(b) statements, there's nothing by that name in the dictionary, so Python looks up the list and finds them in the global dictionary.

Now we get to c+=1 , which is, of course, equivalent to c=c+1 . When Python scans that line, it says "aha, there's a variable named c, I'll put it into my local scope dictionary." Then when it goes looking for a value for c for the c on the right hand side of the assignment, it finds its local variable named c , which has no value yet, and so throws the error.

The statement global c mentioned above simply tells the parser that it uses the c from the global scope and so doesn't need a new one.

The reason it says there's an issue on the line it does is because it is effectively looking for the names before it tries to generate code, and so in some sense doesn't think it's really doing that line yet. I'd argue that is a usability bug, but it's generally a good practice to just learn not to take a compiler's messages too seriously.

If it's any comfort, I spent probably a day digging and experimenting with this same issue before I found something Guido had written about the dictionaries that Explained Everything.

Update, see comments:

It doesn't scan the code twice, but it does scan the code in two phases, lexing and parsing.

Consider how the parse of this line of code works. The lexer reads the source text and breaks it into lexemes, the "smallest components" of the grammar. So when it hits the line

it breaks it up into something like

The parser eventually wants to make this into a parse tree and execute it, but since it's an assignment, before it does, it looks for the name c in the local dictionary, doesn't see it, and inserts it in the dictionary, marking it as uninitialized. In a fully compiled language, it would just go into the symbol table and wait for the parse, but since it WON'T have the luxury of a second pass, the lexer does a little extra work to make life easier later on. Only, then it sees the OPERATOR, sees that the rules say "if you have an operator += the left hand side must have been initialized" and says "whoops!"

The point here is that it hasn't really started the parse of the line yet . This is all happening sort of preparatory to the actual parse, so the line counter hasn't advanced to the next line. Thus when it signals the error, it still thinks its on the previous line.

As I say, you could argue it's a usability bug, but its actually a fairly common thing. Some compilers are more honest about it and say "error on or around line XXX", but this one doesn't.

• 7 Note on implementation details: In CPython, the local scope isn't usually handled as a dict , it's internally just an array ( locals() will populate a dict to return, but changes to it don't create new locals ). The parse phase is finding each assignment to a local and converting from name to position in that array, and using that position whenever the name is referenced. On entry to the function, non-argument locals are initialized to a placeholder, and UnboundLocalError s happen when a variable is read and its associated index still has the placeholder value. –  ShadowRanger Commented Mar 25, 2016 at 19:49
• Python 3.x does not keep local variables in a dictionary. The result of locals() is computed on the fly. This is why the error is called UnboundLocalError in the first place: the local variable exists in the sense that it was reserved ahead of time, when the function was compiled , but hasn't been bound (assigned) yet. This works fundamentally differently from adding something to the global namespace (which is effectively a dictionary), so it wouldn't make sense to report the problem as a generic NameError . –  Karl Knechtel Commented Feb 7, 2023 at 1:08
• Which exactly what I said, modulo the implementation detail of how the locals() are presented. –  Charlie Martin Commented Feb 13, 2023 at 22:18
• "Then when it goes looking for a value for c for the c on the right hand side of the assignment, it finds its local variable named c, which has no value yet, and so throws the error" But if I instead change the line c += 1 as c = 100 , why is the error still present? Python-3.10 –  NameError Commented Oct 26, 2023 at 12:05
• @NameErrorit seems to work as expected for me –  Charlie Martin Commented Nov 5, 2023 at 21:58

Taking a look at the disassembly may clarify what is happening:

As you can see, the bytecode for accessing a is LOAD_FAST , and for b, LOAD_GLOBAL . This is because the compiler has identified that a is assigned to within the function, and classified it as a local variable. The access mechanism for locals is fundamentally different for globals - they are statically assigned an offset in the frame's variables table, meaning lookup is a quick index, rather than the more expensive dict lookup as for globals. Because of this, Python is reading the print a line as "get the value of local variable 'a' held in slot 0, and print it", and when it detects that this variable is still uninitialised, raises an exception.

Python has rather interesting behavior when you try traditional global variable semantics. I don't remember the details, but you can read the value of a variable declared in 'global' scope just fine, but if you want to modify it, you have to use the global keyword. Try changing test() to this:

Also, the reason you are getting this error is because you can also declare a new variable inside that function with the same name as a 'global' one, and it would be completely separate. The interpreter thinks you are trying to make a new variable in this scope called c and modify it all in one operation, which isn't allowed in Python because this new c wasn't initialized.

• Thanks for your response, but I don't think it explains why the error is thrown at line (A), where I'm merely trying to print a variable. The program never gets to line (B) where it is trying to modify an un-initialized variable. –  tba Commented Dec 16, 2008 at 3:42
• 1 Python will read, parse and turn the whole function into internal bytecode before it starts running the program, so the fact that the "turn c to local variable" happens textually after the printing of the value doesn't, as it were, matter. –  Vatine Commented Dec 16, 2008 at 10:10
• 1 Python lets you access global variables in a local scope for reading, but not for writing. This answer has a nice work-around with explanation in comment below... +=1. –  Mark Seagoe Commented Jul 21, 2022 at 12:13

The best example that makes it clear is:

when calling foo() , this also raises UnboundLocalError although we will never reach to line bar=0 , so logically local variable should never be created.

The mystery lies in " Python is an Interpreted Language " and the declaration of the function foo is interpreted as a single statement (i.e. a compound statement), it just interprets it dumbly and creates local and global scopes. So bar is recognized in local scope before execution.

For more examples like this Read this post: http://blog.amir.rachum.com/blog/2013/07/09/python-common-newbie-mistakes-part-2/

This post provides a Complete Description and Analyses of the Python Scoping of variables:

• 1 Python is not any more "interpreted" than Java or C#, and in fact the decision to treat bar as a local variable in this code requires an up-front compilation step. –  Karl Knechtel Commented Sep 9, 2022 at 9:37

Python decides the scope of the variable ahead of time . Unless explicitly overridden using the global or nonlocal (in 3.x) keywords, variables will be recognized as local based on the existence of any operation that would change the binding of a name. That includes ordinary assignments, augmented assignments like += , various less obvious forms of assignment (the for construct, nested functions and classes, import statements...) as well as un binding (using del ). The actual execution of such code is irrelevant.

This is also explained in the documentation .

Contrary to popular belief, Python is not an "interpreted" language in any meaningful sense. (Those are vanishingly rare now.) The reference implementation of Python compiles Python code in much the same way as Java or C#: it is translated into opcodes ("bytecode") for a virtual machine , which is then emulated. Other implementations must also compile the code - so that SyntaxError s can be detected without actually running the code, and in order to implement the "compilation services" portion of the standard library.

How Python determines variable scope

During compilation (whether on the reference implementation or not), Python follows simple rules for decisions about variable scope in a function:

If the function contains a global or nonlocal declaration for a name, that name is treated as referring to the global scope or the first enclosing scope that contains the name, respectively.

Otherwise, if it contains any syntax for changing the binding (either assignment or deletion) of the name, even if the code would not actually change the binding at runtime , the name is local .

Otherwise, it refers to either the first enclosing scope that contains the name, or the global scope otherwise.

Importantly, the scope is resolved at compile time . The generated bytecode will directly indicate where to look. In CPython 3.8 for example, there are separate opcodes LOAD_CONST (constants known at compile time), LOAD_FAST (locals), LOAD_DEREF (implement nonlocal lookup by looking in a closure, which is implemented as a tuple of "cell" objects), LOAD_CLOSURE (look for a local variable in the closure object that was created for a nested function), and LOAD_GLOBAL (look something up in either the global namespace or the builtin namespace).

There is no "default" value for these names . If they haven't been assigned before they're looked up, a NameError occurs. Specifically, for local lookups, UnboundLocalError occurs; this is a subtype of NameError .

Special (and not-special) cases

There are some important considerations here, keeping in mind that the syntax rule is implemented at compile time, with no static analysis :

• It does not matter if the global variable is a builtin function etc., rather than an explicitly created global: def x(): int = int('1') # `int` is local! (Of course, it is a bad idea to shadow builtin names like this anyway, and global cannot help - just like using the same code outside of a function will still cause problems .)
• It does not matter if the code could never be reached: y = 1 def x(): return y # local! if False: y = 0
• It does not matter if the assignment would be optimized into an in-place modification (e.g. extending a list) - conceptually, the value is still assigned, and this is reflected in the bytecode in the reference implementation as a useless reassignment of the name to the same object: y = [] def x(): y += [1] # local, even though it would modify `y` in-place with `global`
• However, it does matter if we do an indexed/slice assignment instead. (This is transformed into a different opcode at compile time, which will in turn call __setitem__ .) y = [0] def x(): print(y) # global now! No error occurs. y[0] = 1
• There are other forms of assignment, e.g. for loops and import s: import sys y = 1 def x(): return y # local! for y in []: pass def z(): print(sys.path) # `sys` is local! import sys
• Another common way to cause problems with import is trying to reuse the module name as a local variable, like so: import random def x(): random = random.choice(['heads', 'tails']) Again, import is assignment, so there is a global variable random . But this global variable is not special ; it can just as easily be shadowed by the local random .
• Deletion is also changing the name binding, e.g.: y = 1 def x(): return y # local! del y

The interested reader, using the reference implementation, is encouraged to inspect each of these examples using the dis standard library module.

Enclosing scopes and the nonlocal keyword (in 3.x)

The problem works the same way, mutatis mutandis , for both global and nonlocal keywords. (Python 2.x does not have nonlocal .) Either way, the keyword is necessary to assign to the variable from the outer scope, but is not necessary to merely look it up , nor to mutate the looked-up object. (Again: += on a list mutates the list, but then also reassigns the name to the same list.)

Special note about globals and builtins

As seen above, Python does not treat any names as being "in builtin scope". Instead, the builtins are a fallback used by global-scope lookups. Assigning to these variables will only ever update the global scope, not the builtin scope. However, in the reference implementation, the builtin scope can be modified: it's represented by a variable in the global namespace named __builtins__ , which holds a module object (the builtins are implemented in C, but made available as a standard library module called builtins , which is pre-imported and assigned to that global name). Curiously, unlike many other built-in objects, this module object can have its attributes modified and del d. (All of this is, to my understanding, supposed to be considered an unreliable implementation detail; but it has worked this way for quite some time now.)

• 1 Simply great. Don't bother with other answers. I still think the UnboundLocalError's message (e.g. local variable 'c' referenced before assignment ) leaves much to be desired. I doubt anybody without intricate knowledge of Python's interpreter inner workings would find it clear –  z33k Commented Mar 11 at 22:57
• @z33k from what I can tell, the usual problem in fixing the bug is in understanding why the variable is local. I don't really see how an error message could explain that, especially since the fact that it's local isn't in itself an error. –  Karl Knechtel Commented Mar 11 at 23:17
• where is c referenced and why is it a problem? Are we talking about references here or is this some other meaning? Why a generic NameError is not sufficient? I know you mention this here but I still don't quite see it judging from what Python tells me and without interpreter knowledge. And I'm not a beginner Python user. –  z33k Commented Mar 12 at 0:10
• 1 It's "reference" in the ordinary English meaning of "try to do something with". It's a subtype because the local namespace works differently, and to try to give more information. Anyway, if you have concrete suggestions for improving how Python does this (and I certainly am not claiming it's perfect), the right place for them is discuss.python.org . –  Karl Knechtel Commented Mar 12 at 1:03

Here are two links that may help

1: docs.python.org/3.1/faq/programming.html?highlight=nonlocal#why-am-i-getting-an-unboundlocalerror-when-the-variable-has-a-value

2: docs.python.org/3.1/faq/programming.html?highlight=nonlocal#how-do-i-write-a-function-with-output-parameters-call-by-reference

link one describes the error UnboundLocalError. Link two can help with with re-writing your test function. Based on link two, the original problem could be rewritten as:

This is not a direct answer to your question, but it is closely related, as it's another gotcha caused by the relationship between augmented assignment and function scopes.

In most cases, you tend to think of augmented assignment ( a += b ) as exactly equivalent to simple assignment ( a = a + b ). It is possible to get into some trouble with this though, in one corner case. Let me explain:

The way Python's simple assignment works means that if a is passed into a function (like func(a) ; note that Python is always pass-by-reference), then a = a + b will not modify the a that is passed in. Instead, it will just modify the local pointer to a .

But if you use a += b , then it is sometimes implemented as:

or sometimes (if the method exists) as:

In the first case (as long as a is not declared global), there are no side-effects outside local scope, as the assignment to a is just a pointer update.

In the second case, a will actually modify itself, so all references to a will point to the modified version. This is demonstrated by the following code:

So the trick is to avoid augmented assignment on function arguments (I try to only use it for local/loop variables). Use simple assignment, and you will be safe from ambiguous behaviour.

The Python interpreter will read a function as a complete unit. I think of it as reading it in two passes, once to gather its closure (the local variables), then again to turn it into byte-code.

As I'm sure you were already aware, any name used on the left of a '=' is implicitly a local variable. More than once I've been caught out by changing a variable access to a += and it's suddenly a different variable.

I also wanted to point out it's not really anything to do with global scope specifically. You get the same behaviour with nested functions.

c+=1 assigns c , python assumes assigned variables are local, but in this case it hasn't been declared locally.

Either use the global or nonlocal keywords.

nonlocal works only in python 3, so if you're using python 2 and don't want to make your variable global, you can use a mutable object:

This issue can also occur when the del keyword is utilized on the variable down the line, after initialization, typically in a loop or a conditional block.

In this case of n = num below, n is a local variable and num is a global variable:

So, there is no error:

But in this case of num = num below, num on the both side are local variables and num on the right side is not defined yet:

So, there is the error below:

UnboundLocalError: local variable 'num' referenced before assignment

In addition, even if removing num = 10 as shown below:

There is the same error below:

So to solve the error above, put global num before num = num as shown below:

Then, the error above is solved as shown below:

Or, define the local variable num = 5 before num = num as shown below:

The best way to reach class variable is directly accesing by class name

• 2 This has nothing to do with the question that was asked. –  Karl Knechtel Commented Feb 7, 2023 at 1:11

You can also get this message if you define a variable with the same name as a method.

For example:

The solution, is to rename method teams() to something else like get_teams() .

Since it is only used locally, the Python message is rather misleading!

You end up with something like this to get around it:

Not the answer you're looking for? Browse other questions tagged python scope global-variables local-variables shadowing or ask your own question .

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