Do not treat a walrus argument as a keyword argument - #242
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davidhalter merged 1 commit intoJul 31, 2026
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f(a := 1, b) was reported as "positional argument follows keyword argument". _ArglistRule inspected only the argument node's first child, so a named expression (name := value) looked identical to a keyword argument (name = value) and set kw_only, making every following positional argument an error. Distinguish the two by the operator child, and add the case to the existing test_valid_namedexpr parametrize list. Closes davidhalter#212
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Closes #212
f(a := 1, b)was reported asSyntaxError: positional argument follows keyword argument._ArglistRulelooked only at the argument node's first child, so a named expression (name := value) was indistinguishable from a keyword argument (name = value) and setkw_only, making every following positional argument an error. Checking the operator child separates the two; thef(**x, y),f(x=2, y)and generator-expression cases are unaffected.The two cases are added to the existing
test_valid_namedexprparametrize list; they fail without the fix and pass with it (full suite: 1988 passed).This change was prepared with AI assistance; the regression test was run locally and fails without the fix.