Haskell Parse error in pattern

Question

isValid :: Position -> Bool
isValid Position(x _) = x
isValid Position(_ y) = y
    | x 'elem' ['a'..'h'] && y 'elem' [1..8] = True
    | otherwise = False

I keep getting this error error: Parse error in pattern: x I am trying to write a function that tells me whether a given poisition is valid or not. Where x is ['a'..'h'] and y is [1..8]

What is the definition of the `Position` type? – dfeuer Jun 01 '17 at 10:31 — dfeuer, Jun 01 '17 at 10:31

score 3 · Answer 1 · answered Jun 01 '17 at 05:37

As explained here: Syntax error on 'mod' Haskell

The syntax for using a named function as an infix operator uses backticks (grave accents, U+0060), not apostrophes:

    | x `elem` ['a'..'h'] && y `elem` [1..8] = True
        ------                 ------
    | otherwise = False

In addition, Position(x _) and Position(_ y) are not valid patterns—you probably intended to use (Position x _) and (Position x y). Note the x, since x is not in scope in the equation you wrote for (Position _ y).

(Position x _) will match all positions, so I suspect you intended:

isValid :: Position -> Bool
isValid (Position x y)
    | x `elem` ['a'..'h'] && y `elem` [1..8] = True
    | otherwise = False

Or more simply:

isValid :: Position -> Bool
isValid (Position x y) = x `elem` ['a'..'h'] && y `elem` [1..8]

Willem Van Onsem · Answer 2 · 2017-06-01T22:51:31.157

I keep getting this error error: Parse error in pattern: x I am trying to write a function that tells me whether a given position is valid or not. Where x is ['a'..'h'] and y is [1..8].

The other answers already discussed what is wrong: you used a guard in the clause where there is no bounded x:

isValid Position(_ y) = y

and furthermore you use quotes instead of backticks with the elem function:

x 'elem' ['a'..'h']

So a rigorous fix would be:

isValid :: Position -> Bool
isValid (Position x y)
    | x `elem` ['a'..'h'] && y `elem` [1..8] = True
    | otherwise = False

Since we actually return the result of the guard, we do not need to use guards and can collapse the guards into one expression:

isValid :: Position -> Bool
isValid (Position x y) = x `elem` ['a'..'h'] && y `elem` [1..8]

Nevertheless since we here work with ranges and the second range are integers, we do not have to use elem on a range, we can use:

isValid :: Position -> Bool
isValid (Position x y) = 'a' <= x && x <= 'h' && 1 <= y && y <= 8

For such small ranges, there will probably not be that much impact on performance, but elem works in O(n) worst case, whereas the two bounds checks work in O(1).

score 2 · Accepted Answer · edited Jun 01 '17 at 08:41

2

In

    | x 'elem' ['a'..'h'] && y 'elem' [1..8] = True

x is unbound. It doesn't appear at all in

isValid Position(_ y) = y

in particular. Also, you probably meant to use `elem`, and not 'elem'.

I am trying to write a function that tells me whether a given poisition is valid or not. Where x is ['a'..'h'] and y is [1..8]

You didn't write the definition of Position, but it seems like this would be something like

data Position = Position Int Int                                                                                                                                     

isValid :: Position -> Bool
isValid (Position x y) = x `elem` ['a'..'h'] && y `elem` [1..8]

(which builds for me).

edited Jun 01 '17 at 08:41

Antal Spector-Zabusky

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answered Jun 01 '17 at 05:37

Ami Tavory

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@Dave You also have another error, in using `'elem'` instead of `\`elem\``, and another one besides that. See updated answer. – Ami Tavory Jun 01 '17 at 05:47
3

@Dave As you seem new to the site (at least by your rep), note that you can accept a single answer, but, independently upvote whatever answers helped you. JonPurdy seems to have answered your question too correctly, so you might want to upvote it as thanks. – Ami Tavory Jun 01 '17 at 05:58

Haskell Parse error in pattern

3 Answers3