On Sonntag, 2. Dezember 2012, 22:46:58, José Romildo Malaquias wrote:

Hello.

What is the exact precedence of the type annotation binary opertor :: ?

It's not really a binary operator. It's part of the syntax, so it has no exact precedence, but since you're asking about it, I presume you're not interested in type declarations foo :: Int -> Double foo = sin . fromIntegral but rather in expression type signatures. The production in the context-free syntax is exp → infixexp :: [context =>] type so the signature is for the entire infix expression: Prelude> toEnum . floor $ 12.7 + toEnum 73 :: Char 'U' hence if it had a precedence, it would be below 0 (the precedence of ($)). But be aware that "The grammar is ambiguous regarding the extent of lambda abstractions, let expressions, and conditionals. The ambiguity is resolved by the meta-rule that each of these constructs extends as far to the right as possible." thus Prelude> (\x -> x + x :: Int -> Int) 2 <interactive>:16:10: No instance for (Num (Int -> Int)) arising from a use of `+' Possible fix: add an instance declaration for (Num (Int -> Int)) In the expression: x + x :: Int -> Int In the expression: \ x -> x + x :: Int -> Int In the expression: (\ x -> x + x :: Int -> Int) 2 the type signature here extends only over the `x + x`, since it is parsed as a part of the lambda abstraction [ \x -> (x + x :: Int -> Int) extends farther to the right than just \x -> x + x] So if you want to give a type signature to a lambda abstraction, you need explicit parentheses: Prelude> ((\x -> x + x) :: Int -> Int) 2 4