[clean-list] Type Expression Grammar question

[Jigang Sun]

>From Clean grammar, 

FunctionType = [Type ->] TypeExpr //A3.1 on page 112 Clean language report

Type = {BrackType}+ 
BrackType = [!] TypeExpr 

TypeExpr = TypeVariable // see A.7 
| PredefinedType 
| (TypeExpr)
PredefinedType = BasicType | ListType | ArrowType
 
ArrowType = "( " Type -> Type ")" //on p43 section 4.6 language report

I cannot see a function type 

myFunc :: Int -> Int -> Int // there are more than one "->"

is accepted. Because in TypeExpr, "->"  must be placed between a pair of parenthesises like (Int -> Int).  

Could the grammar be understood as

FunctionType =  TypeExpr | ArrowType //equivalent to FunctionType = [Type ->] TypeExpr  

SuperTypeExpr=  ArrowType | TypeExpr //used for funcType

TypeExpr = TypeVariable // see A.7 
| PredefinedType 
| (TypeExpr)
PredefinedType = BasicType | ListType |"(" ArrowType ")"
 
ArrowType = Type -> TypeExpr //p43 4.6[1] change from "(" Type -> Type ")"

Here I take FuncType as TypeExp, since a function is a value.

But I still cannot classify ArrowType as TypeExp because 

myFunc :: Int -> Int Int -> Int 

is not be parsed by Clean as

myFunc :: (Int -> Int) Int -> Int

Could anyone explain the related grammar or specify the grammar more clearly?
Thanks.
Jigang





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