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no error when sentinel name does not match variable name #4091

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foo = sentinel('bar') # no error

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see https://peps.python.org/pep-0661/#typing

If the name passed to sentinel() does not match the name the object is assigned to, type checkers should emit an error.

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986ea67d6

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    diagnosticsRelated to reporting of diagnostics.typing semanticstyping-module features, spec compliance, etc

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