Rename Red Knot (#17820)

This commit is contained in:
Micha Reiser 2025-05-03 19:49:15 +02:00 committed by GitHub
parent e6a798b962
commit b51c4f82ea
No known key found for this signature in database
GPG key ID: B5690EEEBB952194
1564 changed files with 1598 additions and 1578 deletions

View file

@ -0,0 +1,114 @@
# Function parameter types
Within a function scope, the declared type of each parameter is its annotated type (or Unknown if
not annotated). The initial inferred type is the union of the declared type with the type of the
default value expression (if any). If both are fully static types, this union should simplify to the
annotated type (since the default value type must be assignable to the annotated type, and for fully
static types this means subtype-of, which simplifies in unions). But if the annotated type is
Unknown or another non-fully-static type, the default value type may still be relevant as lower
bound.
The variadic parameter is a variadic tuple of its annotated type; the variadic-keywords parameter is
a dictionary from strings to its annotated type.
## Parameter kinds
```py
from typing import Literal
def f(a, b: int, c=1, d: int = 2, /, e=3, f: Literal[4] = 4, *args: object, g=5, h: Literal[6] = 6, **kwargs: str):
reveal_type(a) # revealed: Unknown
reveal_type(b) # revealed: int
reveal_type(c) # revealed: Unknown | Literal[1]
reveal_type(d) # revealed: int
reveal_type(e) # revealed: Unknown | Literal[3]
reveal_type(f) # revealed: Literal[4]
reveal_type(g) # revealed: Unknown | Literal[5]
reveal_type(h) # revealed: Literal[6]
# TODO: should be `tuple[object, ...]` (needs generics)
reveal_type(args) # revealed: tuple
reveal_type(kwargs) # revealed: dict[str, str]
```
## Unannotated variadic parameters
...are inferred as tuple of Unknown or dict from string to Unknown.
```py
def g(*args, **kwargs):
# TODO: should be `tuple[Unknown, ...]` (needs generics)
reveal_type(args) # revealed: tuple
reveal_type(kwargs) # revealed: dict[str, Unknown]
```
## Annotation is present but not a fully static type
The default value type should be a lower bound on the inferred type.
```py
from typing import Any
def f(x: Any = 1):
reveal_type(x) # revealed: Any | Literal[1]
```
## Default value type must be assignable to annotated type
The default value type must be assignable to the annotated type. If not, we emit a diagnostic, and
fall back to inferring the annotated type, ignoring the default value type.
```py
# error: [invalid-parameter-default]
def f(x: int = "foo"):
reveal_type(x) # revealed: int
# The check is assignable-to, not subtype-of, so this is fine:
from typing import Any
def g(x: Any = "foo"):
reveal_type(x) # revealed: Any | Literal["foo"]
```
## Stub functions
```toml
[environment]
python-version = "3.12"
```
### In Protocol
```py
from typing import Protocol
class Foo(Protocol):
def x(self, y: bool = ...): ...
def y[T](self, y: T = ...) -> T: ...
class GenericFoo[T](Protocol):
def x(self, y: bool = ...) -> T: ...
```
### In abstract method
```py
from abc import abstractmethod
class Bar:
@abstractmethod
def x(self, y: bool = ...): ...
@abstractmethod
def y[T](self, y: T = ...) -> T: ...
```
### In function overload
```py
from typing import overload
@overload
def x(y: None = ...) -> None: ...
@overload
def x(y: int) -> str: ...
def x(y: int | None = None) -> str | None: ...
```