Haskell中的Lambda抽象模式匹配

Question

I am trying to write a function which outputs all variable names. When used on the example below, the output should be

*Main> used example
     ["a","b","x","y"]

This is what I have written so far...

import Data.Char
import Data.List

type Var = String

data Term =
    Variable Var
  | Lambda   Var  Term
  | Apply    Term Term
--  deriving Show

instance Show Term where
  show = pretty

example :: Term
example = Lambda "a" (Lambda "x" (Apply (Apply (Lambda "y" (Variable "a")) (Variable "x")) (Variable "b")))


pretty :: Term -> String
pretty = f 0
    where
      f i (Variable x) = x
      f i (Lambda x m) = if i /= 0 then "(" ++ s ++ ")" else s where s = "\\" ++ x ++ ". " ++ f 0 m 
      f i (Apply  n m) = if i == 2 then "(" ++ s ++ ")" else s where s = f 1 n ++ " " ++ f 2 m

used :: Term -> [Var]
used (Variable n) = [n]
used (Lambda n t) = "\\" ++ n ++ ". " ++ used t
used (Apply t1 t2) = used t1 ++ used t2

The problem lies in the line used (Lambda nt) = "\\\\" ++ n ++ ". " ++ used t , I get this error message:

list.hs:28:47: error:
    lexical error in string/character literal at character '\n'
   |
28 | used (Lambda n t) = "\" ++ n ++ ". " ++ used t

Why am I getting this complaint?

Answer 1

You are copying and pasting from examples too much. Keep thinking about what the function is supposed to mean as you are writing it.

used (Variable n) = [n]

This is correct: the set of variables used in a term which is just a variable is the variable used. For example, the set of variables used in the term x is just [x] .

used (Apply t1 t2) = used t1 ++ used t2

This is also correct ¹ : the set of variables used in say (xyz) (wxy) is the union of the variables used in xyz and in wxy : that is [x,y,z,w,x,y] .

So now let's consider the case you are struggling with.

used (Lambda n t) = ...

The code you have seems to be attempting to print the the term into a string, which is not what we are trying to do here -- we are trying to find the set of used variables.

Let's consider an example: we have a term like (\\x. xyz) , and we want to find out what its used variables are. Before trying to come up with the general solution, ask yourself what the result should be in this example.

If this function is going to be nicely recursive, can we express your expected result in terms of the used variables of xyz ? How do we transform the set of used variables of xyz into the set of used variables of \\x. xyz \\x. xyz ?

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¹ Though you might get duplicates.