Haskell data type Counts for State Monad

Question

How can I use the following data type for a Monoid instance of it?

data Counts = Counts {
    binds   :: Int,
    returns :: Int,
    gets    :: Int,
    puts    :: Int
} deriving (Eq, Show)

E.g. I figured something like:

mempty  = Counts { 0, 0, 0, 0 }
(Counts { b, r, g, p }) mappend (Counts { b', r', g', p' })  = Counts { (b + b'), (r + r'), (g + g'), (p + p') }

But that gives me a parse error on '0'... Maybe I'm doing it totally wrong and misunderstanding the data type / monoid, but I cannot figure it out. If anyone could help me out it'd be much appreciated!

Best regards, Skyfe.

another slight syntax error is mappend - you have to enclose it in these ``` characters (sadly the same SO uses to make inline code-blocks — Random Dev, Oct 15 '14 at 10:51
@CarstenKönig Aah that's right I totally forgot to do that, thank you — user2999349, Oct 15 '14 at 10:56
Separate to the question you asked, but: you very likely want to use strict fields (replace `Int` with `!Int`) to avoid space leaks. — Michael Snoyman, Oct 15 '14 at 11:18

score 5 · Accepted Answer · answered Oct 15 '14 at 10:41

5

This works:

Counts 0 0 0 0

This works:

Counts {binds = 0, returns = 0, gets = 0, puts = 0}

This does not work:

Counts {0, 0, 0, 0}

You either have to put the brackets and the field names, or you leave the brackets out completely. One or the other.

Other than that, your approach looks sane. You just have your record syntax messed up.

answered Oct 15 '14 at 10:41

MathematicalOrchid

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Aah figured I'd have the syntax messed up, thank you for helping out! – user2999349 Oct 15 '14 at 10:55

Haskell data type Counts for State Monad

1 Answers1