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Combine list splat parameter and arguments assignments

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This commit is contained in:
Rick Winfrey 2017-07-11 15:43:26 -07:00
parent e61251f589
commit a83a6a7e49
1 changed files with 4 additions and 3 deletions
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7
src/Language/Python/Syntax.hs
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@ -154,7 +154,7 @@ expression = await
<|> expressionList <|> expressionList
<|> keywordArgument <|> keywordArgument
<|> keywordIdentifier <|> keywordIdentifier
<|> listSplatParameter <|> listSplat
<|> literal <|> literal
<|> memberAccess <|> memberAccess
<|> notOperator <|> notOperator
@ -175,8 +175,9 @@ typedDefaultParameter = symbol TypedDefaultParameter >>= \ loc -> children (make
makeAnnotation loc identifier' type' value' = makeTerm loc (Type.Annotation (makeAssignment loc identifier' value') type') makeAnnotation loc identifier' type' value' = makeTerm loc (Type.Annotation (makeAssignment loc identifier' value') type')
makeAssignment loc identifier' value' = makeTerm loc (Statement.Assignment identifier' value') makeAssignment loc identifier' value' = makeTerm loc (Statement.Assignment identifier' value')
listSplatParameter :: Assignment listSplat :: Assignment
listSplatParameter = makeTerm <$> symbol ListSplatParameter <*> (Syntax.Identifier <$> source) listSplat = makeTerm <$> symbol ListSplatParameter <*> (Syntax.Identifier <$> source)
<|> makeTerm <$> symbol ListSplatArgument <*> (Syntax.Identifier <$> source)
dictionarySplat :: Assignment dictionarySplat :: Assignment
dictionarySplat = makeTerm <$> symbol DictionarySplatParameter <*> (Syntax.Identifier <$> source) dictionarySplat = makeTerm <$> symbol DictionarySplatParameter <*> (Syntax.Identifier <$> source)