Parsing S Expression

Question

Given the following definitions that make up an S Expression from Prof. Yorgey's course:

data Atom = N Integer | I Ident deriving Show

and

data SExpr = A Atom | Comb [SExpr] deriving Show

What should the full data type be (in Haskell) for the following?

(bar (foo) 3 5 874)

Answer 1

I believe it would be something like

Comb
    [ A (I "bar")
    , Comb
        [ A (I "foo")
        ]
    , A (N 3)
    , A (N 5)
    , A (N 874)
    ]

Whenever you encounter an open parenthesis you would start a new Comb expression, so (foo) is Comb [A (I "foo")] while foo is simply A (I "foo").

Answer 2

I'll assume the Ident type is a String.

• bar as an Atom is I "bar", and as an SExpr is A (I "bar")
• ditto for foo
• (foo) is an SExpr and is constructed as Comb [ A (I "foo") ]
• 3 as an Atom is N 3 and as an SExpr is A (N 3)
• ditto for 5 and 874
• the complete construction of (bar (foo) 3 5 874) (which is an SExpr) is

.

Comb [ A (I "bar")
     , Comb [ A (I "foo") ]
     , A (N 3)
     , A (N 5)
     , A (N 874)
     ]