`str.startswith`

We special-case str.startswith to allow inference of precise Boolean literal types, because those are used in sys.platform checks.

reveal_type("abc".startswith(""))  # revealed: Literal[True]
reveal_type("abc".startswith("a"))  # revealed: Literal[True]
reveal_type("abc".startswith("ab"))  # revealed: Literal[True]
reveal_type("abc".startswith("abc"))  # revealed: Literal[True]

reveal_type("abc".startswith("abcd"))  # revealed: Literal[False]
reveal_type("abc".startswith("bc"))  # revealed: Literal[False]

reveal_type("AbC".startswith(""))  # revealed: Literal[True]
reveal_type("AbC".startswith("A"))  # revealed: Literal[True]
reveal_type("AbC".startswith("Ab"))  # revealed: Literal[True]
reveal_type("AbC".startswith("AbC"))  # revealed: Literal[True]

reveal_type("AbC".startswith("a"))  # revealed: Literal[False]
reveal_type("AbC".startswith("aB"))  # revealed: Literal[False]

reveal_type("".startswith(""))  # revealed: Literal[True]

reveal_type("".startswith(" "))  # revealed: Literal[False]

Make sure that we fall back to bool for more complex cases:

reveal_type("abc".startswith("b", 1))  # revealed: bool
reveal_type("abc".startswith("bc", 1, 3))  # revealed: bool

reveal_type("abc".startswith(("a", "x")))  # revealed: bool

And similiarly, we should still infer bool if the instance or the prefix are not string literals:

from typing_extensions import LiteralString

def _(string_instance: str, literalstring: LiteralString):
    reveal_type(string_instance.startswith("a"))  # revealed: bool
    reveal_type(literalstring.startswith("a"))  # revealed: bool

    reveal_type("a".startswith(string_instance))  # revealed: bool
    reveal_type("a".startswith(literalstring))  # revealed: bool

1.8 KiB Raw Blame History

str.startswith

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Raw Blame History

`str.startswith`