Pylance does not seem to work with implicit type definitions using Protocols and @property

[jaycosaur]

Environment data

• Language Server version: v2020.8.0
• OS and version: macOS Cataline 10.15.5 (19F101)
• Python version: 3.8.2
• VSCode version: 1.47.3

Expected behaviour

Protocols should be interpreted correctly when implicitly defined and using @Property decorator. Though if the @Property method is not used opting for the shorter syntax with a public access variable, no error is produced.

Actual behaviour

Strange error messages appear (with the red underline) stating things like:
"int" is incompatible with "int"Pylance (reportGeneralTypeIssues) or
"str" is incompatible with "str"Pylance (reportGeneralTypeIssues)

Code Snippet / Additional information

from typing import Protocol
from dataclasses import dataclass

class HasAgeProtocol(Protocol):
    @property
    def age(self) -> int:
        ...


@dataclass
class WithNameAndAge:
    age: int
    name: str


someone = WithNameAndAge(age=25, name="Jane")


def print_age(v: HasAgeProtocol):
    return f"{v.age}"


print_age(someone)

This produces the error:
Argument of type "WithNameAndAge" cannot be assigned to parameter "v" of type "HasAgeProtocol" in function "print_age" "age" is an incompatible type "int" is incompatible with "int"Pylance (reportGeneralTypeIssues)

from typing import Protocol
from dataclasses import dataclass

class HasAgeProtocol(Protocol):
    age: int


@dataclass
class WithNameAndAge:
    age: int
    name: str


someone = WithNameAndAge(age=25, name="Jane")


def print_age(v: HasAgeProtocol):
    return f"{v.age}"


print_age(someone)

This works without error.

[erictraut]

This behavior is intended. The protocol you have defined specifies that the class must have a property called age, but your dataclass does not have any such property. A property is an object with different semantics than an instance variable, so they are not interchangeable. For example a property is not settable or deletable unless you provide a setter and a deleter.

If you change your protocol class to the following, it will work as you expect:

class HasAgeProtocol(Protocol):
    age: int

Alternatively, if you define a class that actually defines a property age, it will work:

class WithNameAndAge:
    @property
    def age(self) -> int:
        ...
    @property
    def name(self) -> str:
        ...

[jaycosaur]

Ah apologies @erictraut my misunderstanding here. Of course that makes sense. Thanks!