The type of __default= in dict.pop and dict.get doesn't look right

[superbobry]

dict.pop is currently defined as

typeshed/stdlib/builtins.pyi

Lines 1086 to 1089 in 1851423

	@overload
	def pop(self, __key: _KT) -> _VT: ...
	@overload
	def pop(self, __key: _KT, __default: _VT | _T) -> _VT | _T: ...

Note the union with two type parameters as a type of __default=. This type is ambiguous if dict.pop is called with a value of type _VT. For example, given

d: dict[str, str]
d.pop("foo")

a type checker can choose any value for _T, since str is a subtype of str | _T regardless of what _T is.

It seems the union was introduced in 2fdcd2e, which unfortunately does not seem to give any reasoning for that. I think we should drop it and use _T instead in both dict.pop and dict.get (and similarly for MutableMapping), i.e.

class dict(Generic[_KT, _VT]):
   # ...
   @overload
   def pop(self, __key: _KT, __default: _T) -> _T: ...

[JelleZijlstra]

Yes, I think you're right. Feel free to send a PR!

[superbobry]

Will do!

[hauntsaninja]

I'm a little confused. In your example, why would d.pop("foo") match the def pop(self, __key: _KT, __default: _VT | _T) -> _VT | _T: ... overload (since __default does not have a default value in that overload)?

But I'm fine with anything as long as the following three cases type check:

from typing import assert_type

def f(d: dict[str, int]):
    assert_type(d.pop("key"), int)
    assert_type(d.pop("key", 1), int)
    assert_type(d.pop("key", None), int | None)

[superbobry]

Ooops, sorry. I was meaning to write d.pop("foo", "bar"). I've updated the type to be dict[str, str] to minimize the edits to the original post.

I think all three cases should type check with the proposed definition.

[superbobry]

Okay, another correction. The right signature for seems to be

class dict(Generic[_KT, _VT]):
   # ...
   @overload
   def pop(self, __key: _KT, __default: _T) -> _VT | _T: ...

because a dict can return a value from an existing entry.