Fix type narrowing in lambda expressions (#16407)

ilevkivskyi · JukkaL · commit 8813968abb65 · 2023-11-10T10:02:15.000Z
Fixes #4297 Fix is straightforward: without properly pushing lambda expression on the stack, the previous fix @JukkaL added for nested functions doesn't work for lambdas (it thinks that we are at global scope).
diff --git a/mypy/checkexpr.py b/mypy/checkexpr.py
@@ -5195,7 +5195,8 @@ def visit_lambda_expr(self, e: LambdaExpr) -> Type:
         else:
             # Type context available.
             self.chk.return_types.append(inferred_type.ret_type)
-            self.chk.check_func_item(e, type_override=type_override)
+            with self.chk.tscope.function_scope(e):
+                self.chk.check_func_item(e, type_override=type_override)
             if not self.chk.has_type(e.expr()):
                 # TODO: return expression must be accepted before exiting function scope.
                 self.accept(e.expr(), allow_none_return=True)
diff --git a/test-data/unit/check-inference-context.test b/test-data/unit/check-inference-context.test
@@ -1482,3 +1482,16 @@ b: Any
 i = i if isinstance(i, int) else b
 reveal_type(i)  # N: Revealed type is "Union[Any, builtins.int]"
 [builtins fixtures/isinstance.pyi]
+
+[case testLambdaInferenceUsesNarrowedTypes]
+from typing import Optional, Callable
+
+def f1(key: Callable[[], str]) -> None: ...
+def f2(key: object) -> None: ...
+
+def g(b: Optional[str]) -> None:
+    if b:
+        f1(lambda: reveal_type(b))  # N: Revealed type is "builtins.str"
+        z: Callable[[], str] = lambda: reveal_type(b)  # N: Revealed type is "builtins.str"
+        f2(lambda: reveal_type(b))  # N: Revealed type is "builtins.str"
+        lambda: reveal_type(b)  # N: Revealed type is "builtins.str"
