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Don't unwrap a lambda whose callee contains a call #572

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14 changes: 14 additions & 0 deletions src/fixit/rules/no_redundant_lambda.py
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Original file line number Diff line number Diff line change
Expand Up @@ -34,6 +34,11 @@ class NoRedundantLambda(LintRule):
Valid("lambda x, y: foo(y=x, x=y)"),
Valid("lambda x, y, *z: foo(x, y, z)"),
Valid("lambda x, y, **z: foo(x, y, z)"),
# The callee is re-evaluated on every call of the lambda, but only once
# if the lambda is replaced by it. See issue #509.
Valid("lambda: datetime.now().isoformat()"),
Valid("lambda x: get_handler().handle(x)"),
Valid("lambda x: foo()[0](x)"),
]
INVALID = [
Invalid("lambda: self.func()", expected_replacement="self.func"),
Expand Down Expand Up @@ -72,6 +77,15 @@ def visit_Lambda(self, node: cst.Lambda) -> None:
),
):
call = cst.ensure_type(node.body, cst.Call)

# Unwrapping is only safe while the callee itself is evaluated on
# every call. If it contains a call of its own, that call would be
# evaluated once, at definition time, and its result frozen:
# lambda: datetime.now().isoformat() -> datetime.now().isoformat
# See issue #509.
if m.findall(call.func, m.Call()):
return

full_name = get_full_name_for_node(call)
if full_name is None:
full_name = "function"
Expand Down