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
- 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__()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.
__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
hex(integer) function takes one integer argument and returns a hexadecimal string with prefix
Where to Go From Here?
Enough theory. Let’s get some practice!
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