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Optimise LiteralType.__eq__ and __hash__ #20423

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Optimise LiteralType.__eq__ and __hash__ #20423

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7 changes: 2 additions & 5 deletions mypy/types.py
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Original file line number Diff line number Diff line change
Expand Up @@ -3242,7 +3242,6 @@ def __init__(
super().__init__(line, column)
self.value = value
self.fallback = fallback
self._hash = -1 # Cached hash value

# NOTE: Enum types are always truthy by default, but this can be changed
# in subclasses, so we need to get the truthyness from the Enum
Expand Down Expand Up @@ -3272,13 +3271,11 @@ def accept(self, visitor: TypeVisitor[T]) -> T:
return visitor.visit_literal_type(self)

def __hash__(self) -> int:
if self._hash == -1:
self._hash = hash((self.value, self.fallback))
return self._hash
return hash(self.value)
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@ilevkivskyi ilevkivskyi Dec 17, 2025

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Is this part actually safe? This will make Literal["foo"] and Literal[b"foo"] have same hash (I don't remember why, but we always store the literal bytes value as a string).

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@sterliakov sterliakov Dec 18, 2025

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There's also Literal[0] vs Literal[False], both 0 and False hash to zero IIRC

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@hauntsaninja hauntsaninja Dec 18, 2025

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It's fine for hashes to collide and doesn't affect correctness. You just don't want too many values to collide because then some of your constant time operations become linear.

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@hauntsaninja hauntsaninja Dec 18, 2025

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Only correctness requirement is that values that compare equal have the same hash


def __eq__(self, other: object) -> bool:
if isinstance(other, LiteralType):
return self.fallback == other.fallback and self.value == other.value
return self.value == other.value and self.fallback == other.fallback
else:
return NotImplemented

Expand Down