8000 Ensure we always infer a valid fallback type for lambda callables by Michael0x2a · Pull Request #13576 · python/mypy · GitHub
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Ensure we always infer a valid fallback type for lambda callables #13576

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Merged
merged 1 commit into from
Sep 5, 2022
Merged

Ensure we always infer a valid fallback type for lambda callables #13576

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4 changes: 4 additions & 0 deletions mypy/checkexpr.py
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Original file line number Diff line number Diff line change
Expand Up @@ -4230,6 +4230,10 @@ def infer_lambda_type_using_context(
callable_ctx = get_proper_type(replace_meta_vars(ctx, ErasedType()))
assert isinstance(callable_ctx, CallableType)

# The callable_ctx may have a fallback of builtins.type if the context
# is a constructor -- but this fallback doesn't make sense for lambdas.
callable_ctx = callable_ctx.copy_modified(fallback=self.named_type("builtins.function"))

if callable_ctx.type_guard is not None:
# Lambda's return type cannot be treated as a `TypeGuard`,
# because it is implicit. And `TypeGuard`s must be explicit.
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15 changes: 15 additions & 0 deletions test-data/unit/check-inference.test
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Expand Up @@ -1276,6 +1276,21 @@ class A:
def h(x: Callable[[], int]) -> None:
pass

[case testLambdaJoinWithDynamicConstructor]
from typing import Any, Union

class Wrapper:
def __init__(self, x: Any) -> None: ...

def f(cond: bool) -> Any:
f = Wrapper if cond else lambda x: x
reveal_type(f) # N: Revealed type is "def (x: Any) -> Any"
return f(3)

def g(cond: bool) -> Any:
f = lambda x: x if cond else Wrapper
reveal_type(f) # N: Revealed type is "def (x: Any) -> Any"
return f(3)

-- Boolean operators
-- -----------------
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0