How does the 'infix' work?

Question

I'm playing with the infixr, infixl and infix declarations. I understand how infixr and infixl works:

-- Test expression: 40 +++ 20 +++ 50 +++ 10 * 10  -- infixr 8 +++ -- Calculated as: (40 +++ (20 +++ (50 +++ 10))) * 10. Result: 630. -- infixl 8 +++ -- Calculated as: (((40 +++ 20) +++ 50) +++ 10) * 10. Result: 800.  -- infixr 6 +++ -- Calculated as: 40 +++ (20 +++ (50 +++ (10 * 10))). Result: 75. -- infixl 6 +++ -- Calculated as: ((40 +++ 20) +++ 50) +++ (10 * 10). Result: 125.  (+++) :: Int -> Int -> Int a +++ b = a + (b `div` 2) 

But I don't understand how the infix keyword works. Am I right in thinking that with infix you always need to specify the order with parenthesis? If so, why is the numeric argument necessary considering that brackets have the highest precedence)?

Answer 1

tl;dr

The r and l refer to the associativity, the number you specify refers to the operator precedence. When you don't specify the associativity you get an operator that can be associated only by explicit parenthesis or when the associativity is non-ambiguous.

Our test data structure

Let's use a data structure to define operators on and understand how associativity works:

data Test = Test String deriving (Eq, Show) 

It will contain the string built with the below operators.

Associativity with infixr and infixl

Now let's define right- and left- associative operators:

(>:) :: Test -> Test -> Test (Test a) >: (Test b) = Test $ "(" ++ a ++ " >: " ++ b ++ ")"  (<:) :: Test -> Test -> Test (Test a) <: (Test b) = Test $ "(" ++ a ++ " <: " ++ b ++ ")"  infixr 6 >: infixl 6 <: 

These operator will construct the string of the resulting operator by explicitly adding the parenthesis to our associated terms.

If we test it out we see that it works correctly:

print $ (Test "1") >: (Test "2") >: (Test "4") -- Test "(1 >: (2 >: 4))"  print $ (Test "1") <: (Test "2") <: (Test "4") -- Test "((1 <: 2) <: 4)" 

"Associativity" with infix

An infix declaration does not specify associativity. So what should happen in those cases? Let's see:

(?:) :: Test -> Test -> Test (Test a) ?: (Test b) = Test $ "(" ++ a ++ " ?: " ++ b ++ ")"  infix 6 ?: 

And then let's try it:

print $ (Test "1") ?: (Test "2") ?: (Test "4") 

Woops, we get:

Precedence parsing error cannot mix `?:' [infix 6] and `?:' [infix 6] in the same infix expression

As you can see the language parser noticed that we didn't specify the associativity of the operator and doesn't know what to do.

If we instead remove the last term:

print $ (Test "1") ?: (Test "2") -- Test "(1 ?: 2)" 

Then the compiler doesn't complain.

To fix the original term we would need to explicitly add parenthesis; for example:

print $ (Test "1") ?: ((Test "2") ?: (Test "4")) -- Test "(1 ?: (2 ?: 4))" 

Live demo

Answer 2

Because

1 +++ 2 < 3 

works fine.