3017b3b687
## Summary Inferred and declared types for function parameters, in the function body scope. Fixes #13693. ## Test Plan Added mdtests. --------- Co-authored-by: Micha Reiser <micha@reiser.io> Co-authored-by: Alex Waygood <Alex.Waygood@Gmail.com>
2.5 KiB
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
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
# TODO: should be `dict[str, str]` (needs generics)
reveal_type(kwargs) # revealed: dict
Unannotated variadic parameters
...are inferred as tuple of Unknown or dict from string to Unknown.
def g(*args, **kwargs):
# TODO: should be `tuple[Unknown, ...]` (needs generics)
reveal_type(args) # revealed: tuple
# TODO: should be `dict[str, Unknown]` (needs generics)
reveal_type(kwargs) # revealed: dict
Annotation is present but not a fully static type
The default value type should be a lower bound on the inferred type.
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.
# 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"]