3 ms·
Not quite. It's equivalent to either contains :: (NameClass, QName) -> Bool or contains :: NameClass -> (QName -> Bool) In other words, you can trea
by Nullabillity 11y ago
Not quite. It's equivalent to either
contains :: (NameClass, QName) -> Bool
or
contains :: NameClass -> (QName -> Bool)
In other words, you can treat it as a function taking two arguments, or one function taking one argument and returning another function.
- theseoafs 11y agoIt is equivalent to the second, not the first. The first is a completely distinct type -- though you can certainly write a function with that type, given the original function, by uncurrying it.
- saurik 11y agoFWIW, I think the person you are responding to is using the non-Haskell requested mental syntax used by the person who asked the original question, in order so they better understand the semantics (even if at a cost of confusing them for later development in Haskell).
- Nullabillity 11y agoYup. They're distinct types, but mentally they are the same (and the standard library has functions to convert between the two forms).