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Rename Red Knot (#17820)
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# Function parameter types
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Within a function scope, the declared type of each parameter is its annotated type (or Unknown if
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not annotated). The initial inferred type is the union of the declared type with the type of the
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default value expression (if any). If both are fully static types, this union should simplify to the
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annotated type (since the default value type must be assignable to the annotated type, and for fully
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static types this means subtype-of, which simplifies in unions). But if the annotated type is
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Unknown or another non-fully-static type, the default value type may still be relevant as lower
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bound.
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The variadic parameter is a variadic tuple of its annotated type; the variadic-keywords parameter is
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a dictionary from strings to its annotated type.
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## Parameter kinds
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```py
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from typing import Literal
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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):
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reveal_type(a) # revealed: Unknown
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reveal_type(b) # revealed: int
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reveal_type(c) # revealed: Unknown | Literal[1]
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reveal_type(d) # revealed: int
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reveal_type(e) # revealed: Unknown | Literal[3]
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reveal_type(f) # revealed: Literal[4]
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reveal_type(g) # revealed: Unknown | Literal[5]
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reveal_type(h) # revealed: Literal[6]
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# TODO: should be `tuple[object, ...]` (needs generics)
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reveal_type(args) # revealed: tuple
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reveal_type(kwargs) # revealed: dict[str, str]
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```
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## Unannotated variadic parameters
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...are inferred as tuple of Unknown or dict from string to Unknown.
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```py
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def g(*args, **kwargs):
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# TODO: should be `tuple[Unknown, ...]` (needs generics)
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reveal_type(args) # revealed: tuple
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reveal_type(kwargs) # revealed: dict[str, Unknown]
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```
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## Annotation is present but not a fully static type
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The default value type should be a lower bound on the inferred type.
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```py
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from typing import Any
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def f(x: Any = 1):
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reveal_type(x) # revealed: Any | Literal[1]
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```
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## Default value type must be assignable to annotated type
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The default value type must be assignable to the annotated type. If not, we emit a diagnostic, and
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fall back to inferring the annotated type, ignoring the default value type.
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```py
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# error: [invalid-parameter-default]
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def f(x: int = "foo"):
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reveal_type(x) # revealed: int
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# The check is assignable-to, not subtype-of, so this is fine:
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from typing import Any
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def g(x: Any = "foo"):
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reveal_type(x) # revealed: Any | Literal["foo"]
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```
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## Stub functions
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```toml
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[environment]
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python-version = "3.12"
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```
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### In Protocol
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```py
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from typing import Protocol
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class Foo(Protocol):
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def x(self, y: bool = ...): ...
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def y[T](self, y: T = ...) -> T: ...
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class GenericFoo[T](Protocol):
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def x(self, y: bool = ...) -> T: ...
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```
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### In abstract method
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```py
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from abc import abstractmethod
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class Bar:
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@abstractmethod
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def x(self, y: bool = ...): ...
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@abstractmethod
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def y[T](self, y: T = ...) -> T: ...
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```
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### In function overload
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```py
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from typing import overload
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@overload
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def x(y: None = ...) -> None: ...
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@overload
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def x(y: int) -> str: ...
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def x(y: int | None = None) -> str | None: ...
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```
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