In this article, we are going to learn what invalid tokens are in Python, how it shows an invalid token in python 2.x and python 3.x compiler. We will also learn how to fix this type of invalid token error in Python. Assigning an octal number for python 2.x and python 3.x is different from each other. Let us start.
Imagine that you are going to develop a python project for an educational center. On that, you need to store the roll number for the student like 01. At the same time, we try to store roll number =01. It will show an error like an invalid token. So how to fix it? In python 2.x compiler, it will store till 07 without any errors. But while we go for 08, it will show an error as an invalid token. In python 3.x, we have to store the octal numbers as 0o1. It doesn’t show any errors till we try to 0o7.
What are tokens in python?
Each line in Python is broken down into different essential lexical components. These components can be keywords, identifiers, operators, delimiters, or other types of features. These essential components are called tokens. By default, tokens are separated by whitespace in each logical line.
For example, consider a logical line in Python –
x_variable = 1
Here, ‘x_variable’, ‘=’, ‘1’ are tokens.
These tokens are then parsed and identified.
Which tokens are parsed as Integer literals?
Integer literals are described by the following lexical definitions:
integer | decimal integer/binary integer/Octal integer/Hexadecimal integer |
decimal integer | nonzero digit([“_”] digit)* | “0”+ ([“_”] “0”)* |
binary integer | 0 (b/B) ([ ]binarydigit) |
octal integer | 0 (o/O) ([ ] octaldigit) |
hexadecimal integer | 0(x/X) ([ ] hexadecimaldigiti) |
binary digit | 0 and 1 |
octal digit | 0 to 7 |
hexadecimal digit | 0 to 15 |
We can declare any number of integer literals we want. There is no limit. It will ignore the underscore to determine the numeric value. Underscores are used to group digits for enhanced readability.
Invalid tokens in python 2.x
In python 2.x interpreter, the prefix 0 indicates that the given number is Octal. We already know that the Octal number is valid only from 0 to 7. From 8, it is not considered an octal number. What if we try to print as 08 in Python 2x interpreter? It is not possible if we try to implement this. It shows an error as an invalid token.
What are Invalid Tokens Errors and how to fix them?
Let us see some of the invalid token errors in python2.x error.
Example 1: Error
>>>09
SyntaxError: invalid token
Fix the error
The above-given value is not octal, so we can’t add prefixes as 0 for the value 9. If we need to fix this error, we have to follow the below-given code:
>>>0x9
9
The value is hexadecimal, so we are declaring it as 0x9. Now it won’t declare any type of error.
Either we can declare it as a hexadecimal value or a string value.
>>>"09"
9
Now it is declared as a string so that it won’t show any errors.
Example 2: Declaring as datatype for invalid tokens
Now we will try to declare the value using int data type.
>>>int(09)
SyntaxError: invalid token
Fix the errors
>>>int("09")
9
Example 3: Using dictionaries
a={"test":09}
print(a)
Output
SyntaxError: invalid token
How to fix this?
a={"test":int("09")}
print(a)
Output
{'test': 9}
Invalid Tokens in Python 3.x
In python 3.x, there is a different way to represent an octal number. We will use 0o1 to represent an octal number in the version of python 3.x. The numbers from 0o1 to 0o7 won’t show any errors. It just prints the value of the element.
Example 1: Error
>>> 0o9
SyntaxError: invalid digit '9' in octal literal
Here we have declared 9 as an octal number. So it shows an invalid token error. To overcome this, we need t give the value 9 as a hexadecimal value.
>>> 0x9
9
Or we can use it as a string instead of a hexadecimal value.
>>> "0o9"
'0o9'
Example 2: Declaring as datatype for invalid tokens
Now we will try to declare the value using int data type.
>>> int(0o9)
SyntaxError: invalid digit '9' in octal literal
Fix the errors
>>> int("09")
9
Example 3: Using dictionaries
>>> a={"Python":"01"}
>>> a
{'Python': '01'}
>>> a={"Python":int("01")}
>>> a
{'Python': 1}
Practical Examples
Let us imagine that we are developing a project for a hospital. Here we have to store like, 01 at that time it shows an error. Let us learn about this in a detailed manner by implementing it in a python program.
>>> medical_records["Jack"] = {"Age": 52, "Sex": "Male", "BMI": 18.7, "Children": 02, "Smoker": "Non-smoker", "Insurance_cost": 6420.0}
SyntaxError: leading zeros in decimal integer literals are not permitted; use an 0o prefix for octal integers
This is the standard error that we face while we need to note. To overcome this error, either we need to save integer 2 as “02”, 2, or int(“02”).
To display the date, year, month, day, we need to give us:
>>> a = (2016,04,03)
SyntaxError: leading zeros in decimal integer literals are not permitted; use an 0o prefix for octal integers
>>> a=("2019","07","04")
>>> a
('2019', '07', '04')
FAQs Related to Invalid tokens in Python
In python 2.x interpreter, the prefix 0 indicates that the given number is Octal.
The numbers from 00 to 07 are the valid tokens since it is an octal number. When we go for 08, it shows a result as invalid tokens.
The prefix 0o is useful to represent an octal number in python 3.x
Conclusion
In this article, we have learned about Invalid tokens in python 2.x and python 3.x compiler. We have seen some of the possible invalid tokens and solutions for those invalid tokens in Python. This is the primary article that every Python beginner has to learn about it. These are the necessary things that you have to know as a python programmer.
In case of any doubts, kindly let us know in the comment section. Learn Python with us!
Reference:
Lexical Analysis for Integer Literals – Python Official Documentation