No, Python __hex__() Does Not Exist. Do This Instead!

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The Problem

TypeError: '...' object cannot be interpreted as an integer

If you’re reading this article, chances are that you have been thinking something along those lines:

  • Given a custom class My_Class. You want to override the behavior of the built-in hex(x) function in Python when calling it on a My_Class object x.
  • You know about the feature of Dunder Methods to overwrite the custom behavior of built-in functions.
  • You conclude that the dunder method for hex() is __hex__() because, after all, this is how it is supposed to work, right?
  • Wrong! If you try to do it, you get the following error.

The Wrong Example – Don’t!

class My_Class:
    def __hex__(self): # WRONG!
        return 42


x = My_Class()
print(hex(x))

The output indicates that something went wrong:

Traceback (most recent call last):
  File "C:\Users\xcent\Desktop\code.py", line 7, in <module>
    print(hex(x))
TypeError: 'My_Class' object cannot be interpreted as an integer

What To Do?

How can you resolve the TypeError: 'My_Class' object cannot be interpreted as an integer?

To resolve the TypeError when trying to override the behavior of the hex() function using the __hex__() method is to forget about the __hex__() method and use the __index__() method instead.

The __index__() method returns an integer value associated with a given object. The integer is then automatically converted to a hexadecimal number.

The Right Example – Do!

Here’s how this works in a simple example:

class My_Class:
    def __index__(self): # Correct!
        return 42


x = My_Class()
print(hex(x))
# 0x2a

In case you’re wondering about the output 0x2a instead of what you’ve defined, i.e., 42, the integer 42 is converted to the hexadecimal number 0x2a. See here:

>>> hex(42)
'0x2a'

Background hex() – Not a Magic Trick

Python’s built-in hex(integer) function takes one integer argument and returns a hexadecimal string with prefix "0x".

Where to Go From Here?

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