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[ty] don't union in default type for annotated parameters (#21208)

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Carl Meyer 2025-11-02 18:21:54 -05:00 • committed by GitHub
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15
crates/ty_python_semantic/resources/mdtest/function/parameters.md
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@ -1,12 +1,9 @@
# 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.
not annotated). The initial inferred type is the annotated type of the parameter, if any. If there
is no annotation, it is the union of `Unknown` with the type of the default value expression (if
any).
The variadic parameter is a variadic tuple of its annotated type; the variadic-keywords parameter is
a dictionary from strings to its annotated type.
@ -41,13 +38,13 @@ def g(*args, **kwargs):
## Annotation is present but not a fully static type
The default value type should be a lower bound on the inferred type.
If there is an annotation, we respect it fully and don't union in the default value type.
```py
from typing import Any
def f(x: Any = 1):
reveal_type(x) # revealed: Any | Literal[1]
reveal_type(x) # revealed: Any
```
## Default value type must be assignable to annotated type
@ -64,7 +61,7 @@ def f(x: int = "foo"):
from typing import Any
def g(x: Any = "foo"):
reveal_type(x) # revealed: Any | Literal["foo"]
reveal_type(x) # revealed: Any
```
## Stub functions