mirror of https://github.com/astral-sh/ruff.git synced 2025-09-28 21:05:08 +00:00

Dhruv Manilawala a69f6240cc

[red-knot] Infer lambda return type as Unknown (#16695 )

## Summary

Part of #15382

This PR infers the return type `lambda` expression as `Unknown`. In the
future, it would be more useful to infer the expression type considering
the surrounding context (#16696).

## Test Plan

Update existing test cases from `@todo` to the (verified) return type.

2025-03-18 22:48:10 +05:30

2.2 KiB

Raw Blame History

`lambda` expression

No parameters

lambda expressions can be defined without any parameters.

reveal_type(lambda: 1)  # revealed: () -> Unknown

# error: [unresolved-reference]
reveal_type(lambda: a)  # revealed: () -> Unknown

With parameters

Unlike parameters in function definition, the parameters in a lambda expression cannot be annotated.

reveal_type(lambda a: a)  # revealed: (a) -> Unknown
reveal_type(lambda a, b: a + b)  # revealed: (a, b) -> Unknown

But, it can have default values:

reveal_type(lambda a=1: a)  # revealed: (a=Literal[1]) -> Unknown
reveal_type(lambda a, b=2: a)  # revealed: (a, b=Literal[2]) -> Unknown

And, positional-only parameters:

reveal_type(lambda a, b, /, c: c)  # revealed: (a, b, /, c) -> Unknown

And, keyword-only parameters:

reveal_type(lambda a, *, b=2, c: b)  # revealed: (a, *, b=Literal[2], c) -> Unknown

And, variadic parameter:

reveal_type(lambda *args: args)  # revealed: (*args) -> Unknown

And, keyword-varidic parameter:

reveal_type(lambda **kwargs: kwargs)  # revealed: (**kwargs) -> Unknown

Mixing all of them together:

# revealed: (a, b, /, c=Literal[True], *args, *, d=Literal["default"], e=Literal[5], **kwargs) -> Unknown
reveal_type(lambda a, b, /, c=True, *args, d="default", e=5, **kwargs: None)

Parameter type

In addition to correctly inferring the lambda expression, the parameters should also be inferred correctly.

Using a parameter with no default value:

lambda x: reveal_type(x)  # revealed: Unknown

Using a parameter with default value:

lambda x=1: reveal_type(x)  # revealed: Unknown | Literal[1]

Using a variadic paramter:

# TODO: should be `tuple[Unknown, ...]` (needs generics)
lambda *args: reveal_type(args)  # revealed: tuple

Using a keyword-varidic parameter:

# TODO: should be `dict[str, Unknown]` (needs generics)
lambda **kwargs: reveal_type(kwargs)  # revealed: dict

Nested `lambda` expressions

Here, a lambda expression is used as the default value for a parameter in another lambda expression.

reveal_type(lambda a=lambda x, y: 0: 2)  # revealed: (a=(x, y) -> Unknown) -> Unknown

2.2 KiB Raw Blame History

lambda expression