Python, `pathlib.Path.open()` method and built-in function `open()` don't return instances of `typing.IO`

Question

I've read some other answers, and they seem contradictory to what happens IRL.

Try the following:

from typing import IO
from pathlib import Path

example_path = Path(r"D:\Example.txt")
with example_path.open("r") as f:
    print(isinstance(f, IO))
with open(example_path) as f:
    print(isinstance(f, IO))

It will print:

False
False

That's baffling, because according to documentation, it seems to me that one of the very purposes of typing.IO is for type-checking an object for IO type, and this would include the use of isinstance(). Why doesn't it work? Doesn't that defeat the very purpose? What am I missing here?

InSync · Accepted Answer · 2024-07-24 04:56:27Z

2

typing.IO and its companions are for type hinting only. At runtime they are defined as abstract generic classes, and the objects returned by open() are not instances of these classes:

class IO(Generic[AnyStr]):
    @property
    @abstractmethod
    def mode(self) -> str: pass

    @property
    @abstractmethod
    def name(self) -> str: pass

    ...

In fact, the objects in question are created in C (for CPython):

>>> type(f)
<class '_io.TextIOWrapper'>
>>> type(f).__bases__
(<class '_io._TextIOBase'>,)

Theses "private" _io types actually map to corresponding io. types, which you can use with isinstance():

>>> import io
>>> isinstance(f, io.TextIOWrapper)
True
>>> isinstance(f, io.TextIOWrapper)
True

The runtime details are covered in-depth at Check if object is file-like in Python.

answered Jul 24, 2024 at 4:56

InSync

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