Incorrect unreachable code detection after return in `contextlib.suppress`

[jacobbogdanov]

Describe the bug
if there is a return statement in a with contextlib.suppress(Exception) block, then all the code after is grayed out as unreachable.

To Reproduce
See the code example below

Expected behavior
Code after a return in a contextlib.suppress is not marked as unreachable.

Screenshots or Code

import contextlib
from typing import NoReturn

def func() -> int:
    with contextlib.suppress(KeyError):
        return func2()

    return 5  # this line is grayed out

def func2() -> NoReturn:
    raise KeyError

VS Code extension or command-line
pyright vscode extension version 1.1.45

[erictraut]

I think this behavior is correct. The call to func2 always generates an exception, so control flow will never continue after it is called. The return 5 statement will never be executed.

Are you assuming that the context manager is somehow catching and suppressing the raised exception? If so, a type checker would have no way of knowing this.

[jacobbogdanov]

From the docs: https://docs.python.org/3/library/contextlib.html#contextlib.suppress

Return a context manager that suppresses any of the specified exceptions if they occur in the body of a with statement and then resumes execution with the first statement following the end of the with statement.

(emphasis mine.)

So while func2 always raises an exception, in this example calling func will return 5.

Here is the discussion about mypys --warn-unreachable flag with regards to contextlib.suppress: python/mypy#7214

[erictraut]

As per discussion in this PR thread (#771), I'm going to close this for now. Fixing it will add significant complexity to the code, and I'm reluctant to fix it given that 1) assuming behavior of the context manager based on its __exit__ method return type is not part of the Python spec, and 2) there's already a well-defined way in the language to catch exceptions (namely, try/except blocks).