Making a ternary logic table, and I would like to make my own function for an operator that I'll call <=>.
So, for example, I want to do this, but that isn't right. what's the correct way to do this?
data Ternary = T | F | M
deriving (Eq, Show, Ord)
<=> :: Ternary -> Ternary -> Ternary
<=> T F = F
<=> T T = T
<=> T M = M
<=> F F = T
<=> F T = F
<=> F M = M
<=> M F = M
<=> M T = M
<=> M M = T
Just add parentheses around your operator:
(<=>) :: Ternary -> Ternary -> Ternary
(<=>) T F = F
(<=>) T T = T
(<=>) T M = M
(<=>) F F = T
(<=>) F T = F
(<=>) F M = M
(<=>) M F = M
(<=>) M T = M
(<=>) M M = T
This turns it from infix form to prefix form. Alternatively, you can just use infix in the definition:
(<=>) :: Ternary -> Ternary -> Ternary
T <=> F = F
T <=> T = T
T <=> M = M
F <=> F = T
F <=> T = F
F <=> M = M
M <=> F = M
M <=> T = M
M <=> M = T
: cons operator? The : cons operator has this special feature where it considers everything on the right hand side to be a list. I've been trying to recreate the : operator, but it always requires parentheses on the right hand side.
infix as well?
data List a = Nil | a :- List a and then the crucial part: infixr 5 :-. Doesn't have to be 5, but that is the precedence of lists, just has to be infixr and NOT infixl or infix. infixl 9 is the defaultFunction names with symbols have different syntax than those without:
-- Works:
(<^>) :: Int -> Int -> Int
a <^> b = a + b
-- Doesn't work:
{-
<^> :: Int -> Int -> Int
<^> a b = a + b
-}
-- Works:
letters :: Int -> Int -> Int
letters a b = a + b
-- Doesn't work:
{-
(letters) :: Int -> Int -> Int
a letters b = a + b
-}
I promise, though - Haskell is well worth learning the complex rules.
You can simplify (line-wise) the definition as follows:
(<=>) :: Ternary -> Ternary -> Ternary
T <=> T = T
F <=> F = T
M <=> M = T
M <=> _ = M
_ <=> M = M
_ <=> _ = F
(line-wise) in there. Clarity is debatable though. I can see the code better because I am forced to deduce what it actually does as opposed to looking at a raw tabulated definition. But that's me.M <=> M being T part.Since you have Eq and Ord, you can do the following:
data Ternary = T | F | M
deriving (Eq, Show, Ord)
(<=>) :: Ternary -> Ternary -> Ternary
x <=> y = if x == y then T else max x y
If you do happen to change it so that M <=> M == M, then you can do the following:
data Ternary = M | T | F
deriving (Eq, Show, Ord, Enum)
(<=>) :: Ternary -> Ternary -> Ternary
x <=> y = fromEnum $ rem (toEnum x * toEnum y) 3
M <=> Mshould beMrather thanT. But that depends on your "Maybe" semantics.infixl,infixr...