unboundlocalerror local variable 'z' referenced before assignment

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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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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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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UnboundLocalError: cannot access local variable 'X' where it is not associated with a value

Last updated: Apr 10, 2024 Reading time · 6 min

# UnboundLocalError: cannot access local variable 'X' where it is not associated with a value

The Python "UnboundLocalError: cannot access local variable 'X' where it is not associated with a value" occurs for multiple reasons:

• Referencing a local variable before assigning a value to it in a function.
• Only assigning a value to a variable if a certain condition is met.
• Only assigning a value to a variable in a try/except statement.

To solve the error, mark the variable as global before referencing it in the function or make sure to assign a value to the variable regardless if a condition is met.

Here is an example of how the error occurs.

We reference the country variable in the function before assigning a value to it.

The local country variable shadows the country variable that is declared in the global scope.

# Mark the variable as global before accessing it within the function

To solve the error, mark the variable as global before accessing it within your function.

When we mark the variable as global , we tell to the interpreter to look for it in the global scope.

The first call to the function reassigns the global country variable, setting it to "Germany".

Once we call the function the second time and print the country variable, it's already set to "Germany".

# Local variables shadow global ones with the same name

You can reference the country global variable from within the function, however, if you assign a value to the variable, the local variable shadows the global one.

If you assign a value to the variable within the function without marking it as global , the assignment only applies to the function's local scope.

The country variable in the function is set to 'Germany' and the country variable outside the function has a value of 'Austria'.

This behavior is very confusing and should generally be avoided.

It is much better to pick different names and avoid clashes.

# Assigning values to local variables from an outer function

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.

If you get the error "Local variable referenced before assignment", check out the following article .

# Returning a value that is used to reassign the global variable

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 assign to the country 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 country 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 .

# Only assigning a value to a variable is a condition is met

Another common cause of the error is assigning a value to a variable only if a condition is met and trying to access it after.

The if statement checks if 5 is greater than 10 .

The condition is never met, so the country variable never gets set.

When we try to access the variable after the if block, we get the error because the variable was never associated with a value.

To solve the error, set the variable to a default value before the if block.

We initialized the country variable to None outside the if block.

We used a default value of None , but you can also use an empty string "" or 0 as the default value, depending on your use case.

The variable is set even if the if condition isn't met, so the error is never raised.

# Only assigning a value to a variable in a try/except statement

Another cause of the error is only assigning a value to a variable in a try/except statement.

Notice that an error is raised before the country variable is declared in the try block.

The variable is never given a value, so when we try to access it in the return statement , an UnboundLocalError occurs.

To solve the error, set the variable to an initial value before the try statement.

We initialized the country variable to None before the try block.

We used None as the default value, but you can also set the variable to an empty string "" or 0 depending on your use case.

Even if an error is raised in the try block, the variable is still associated with a value because we've initialized it prior.

# Additional Resources

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

• Python: Convert string with comma separator and dot to float
• Format number with comma as thousands separator in Python
• Not all arguments converted during string formatting (Python)
• Python: locale.Error: unsupported locale setting [Solved]

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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:

Watch a video course Python - The Practical Guide

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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What is UnboundLocalError: local variable referenced before assignment?

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

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

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

An example of Local variable referenced before assignment

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

Firstly, we start by declaring two variables:

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

Finally, we call our function:

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

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

Here is an error!

The Solution of Local variable referenced before assignment

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

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

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

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

Let’s try to run our code again:

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

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

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

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

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

Our code returns: B.

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

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

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Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs when you try to use a local variable before it has been assigned a value. This is a general programming concept describing the situation 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.

In Python, the compiler might throw the exact error: “UnboundLocalError: cannot access local variable ‘x’ where it is not associated with a value”

Here’s an example to illustrate this error:

In this example, you would encounter the above 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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having a problem with error code: UnboundLocalError: local variable 'player' referenced before assignment

hi hoping someone here can actually help me. I asked this question on stack overflow and they closed it and redirected to similar previously asked questions. none of which actually helped me to understand how to correct the error code.

I am following a video tutorial series for learning to code with python/pygame. and i have checked my coding aginst the source material and they are the same, because i typed it as was told to in the video.

here's the full tracebackback including the error:

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 160, in <module>

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 152, in main

GameInterface(num_player=1, screen=screen)

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 82, in GameInterface

if player.cooling_time > 0:

UnboundLocalError: local variable 'player' referenced before assignment.

here's the actual code:

if player.cooling_time > 0: player.cooling_time -= 1

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

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在Python编程中，UnboundLocalError是一个运行时错误，它发生在尝试访问一个在当前作用域内未被绑定（即未被赋值）的局部变量时。 错误信息UnboundLocalError: local variable ‘xxx’ referenced before assignment指出变量xxx在赋值之前就被引用了。 这种情况通常发生在函数内部，尤其是在使用循环或条件语句时，变量的赋值逻辑可能因为某些条件未满足而未能执行，导致在后续的代码中访问了未初始化的变量。

我们来看看粉丝跟我说的具体的报错情况：

运行后会显示报错：UnboundLocalError: local variable ‘xxx’ referenced before assignment

把变量声明称global，global sum_score。

• 条件语句中未初始化变量
• 循环中变量初始化位置错误
• 循环的退出条件导致变量未初始化
• 确保变量在使用前被初始化
• 调整循环中变量的作用域
• 检查循环退出条件，确保变量被初始化
• 明确变量作用域：理解Python中变量的作用域，确保在变量的作用域内使用前已经初始化。
• 使用初始化值：为变量提供一个初始值，特别是在不确定变量是否会被赋值的情况下。
• 条件语句的使用：在条件语句中使用变量前，确保变量已经在所有分支中被初始化。
• 循环逻辑检查：在循环中使用变量前，确保循环的逻辑允许变量被正确初始化。
• 代码审查：定期进行代码审查，检查变量的使用是否符合预期，特别是变量初始化的逻辑。
• 编写测试：编写单元测试来验证函数或方法在所有预期的使用情况下都能正确处理变量初始化。

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UnboundLocalError: local variable 'x' referenced before assignment

I am trying to execute this code,

and I am constantly receiving this error,

I have tried global and nonlocal, but it does not work. Since I am not gettin any input from outside the function, so I anyways would not require global or nonlocal.

• unboundlocalerror

• 1 You might want to make sure you're not shoving an empty geometry object into that function. if the for loop never executes, then yeah, there won't be an x. In fact, are those if statements supposed to be executed in the loop? If so, you need to indent them properly. Otherwise, you're only checking the last point (if there are any). –  Lou Commented Apr 30, 2012 at 18:01

If you meant to execute the lines following x,y,z = geometry.GetPoint(i) within the for loop, they must be indented an additional 4 spaces. Remember in Python, indentation is part of the syntax .

• 1 If you run code in IDLE before hand it will spot indentation and syntax errors and point out where they are. –  James Milner Commented Apr 30, 2012 at 20:07
• 1 In this case it's not a syntax error but a logic error and would not be caught by the IDE. –  blah238 Commented Apr 30, 2012 at 20:29

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UnboundLocalError: local variable 'L' referenced before assignment Python [duplicate]

when trying to compile the code below I get this error

Can someone explain why ? Isn't a global variable assigned before anything else?

My Python version is 2.7.3

• 1 Note that UnboundLocalError is a run-time exception, not a compiler exception. –  Martijn Pieters ♦ Commented Jan 30, 2014 at 12:43

3 Answers 3

The minimal code to reproduce your bug is

This is happening for a number of reasons

• First - because in python we have mutable and immutable classes. Ints are immutable, that is when you write x+=1 you actually create another object (which is not true for certain ints due to optimisations CPython does). What actually happens is x = x + 1.
• Second - because python compiler checks every assignment made inside a scope and makes every variable assigned inside that scope local to it.
• So as you see when you try to increment x compiler has to access a variable that's local to that scope, but was never assigned a value before.

If you're using python2 - you only have the option to declare variable global . But this way you would be unable to get a variable from an in-between function like

In python3 you have nonlocal keyword to address this situation.

Also I would advise you to avoid using globals. Also there is a collection.Counter class that might be useful to you.

Further reading: python docs

Isn't a global variable assigned before anything else?

Yes, but that's completely irrelevant. The compiler sees an assignment within the function and marks the name as being in the local scope. You need to use the global keyword at the beginning of the function to tell the compiler that the name should be in the global scope instead.

• Or, better, redesign to not use a global variable at all, since it's a really terrible design. –  Wooble Commented Jan 30, 2014 at 12:42

You are mixing tabs and spaces; don't do that.

Run your script with python -tt yourscript.py and fix all errors that finds.

Then configure your editor to stick to only spaces for indentation; using 4 spaces per indent is the recommended style by the Python Style Guide .

Next, you are trying to increment the global L here:

without declaring it a global. Add global L in that function. Assignment to a name inside a function marks such a name as a local, unless you specifically tell Python it is not.

• @Wooble: Yup, you are correct; I missed the L+=1 in the function there. The indentation in the post was terrible and there were actual tabs on the later lines with L and a for loop. –  Martijn Pieters ♦ Commented Jan 30, 2014 at 12:42
• Thank you for that. I fixed 2 mistakes but didn't solve my problem. –  lvi Commented Jan 30, 2014 at 12:42

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