pyright not warning about mismatch Generator with yield from?

[Mikaël Capelle (Holt59)]

The following code does not generate an error from pyright:

def fn(n: int) -> Generator[int, str, None]:
    yield from range(n)

My understanding is that the actual return type of fn would be Iterator[int] (reveal_type(iter(range(6))) shows Iterator[int]) which mismatches the declared return type. This generates error at runtime if anyone tries to send something to the generator.

I am wondering if this is by design or an overlook? If by design, what is the reasoning?

Thanks,

[Rich Chiodo (rchiodo)]

AFAIK, that's correct. yield causes a Generator, which is a subclass of Iterator. You can check that with the runtime:

def fn(n: int):
    yield from range(n)

x = fn(4)
print(type(x))

That prints this:

<class 'generator'>

[Mikaël Capelle (Holt59)]

Even if yield causes a Generator, it should be Generator[int, None, None], which is incompatible with Generator[int, str, None]. If you consider the following:

from typing import Generator

reveal_type(iter(range(4)))  # Iterator[int]

def fn1(n: int) -> Generator[int, None, None]:
    yield from range(n)

reveal_type(fn1(4))  # Generator[int, None, None]

def fn2(n: int) -> Generator[int, str, None]:
    yield from range(n)

reveal_type(fn2(4))  # Generator[int, str, None]

def fn3(n: int) -> Generator[int, str, None]:
    yield from fn1(n)  # error here due to incompatible return type

reveal_type(fn3(4))  # Generator[int, str, None]

So in fn2, the "return type" of yield from range(n) is (AFAIK) Iterator[int] which is a mismatch from Generator[int, str, None] (it would be okay the other way around).