3 ms·
Maybe I'm misunderstanding what you're saying, but how on Earth would a function f :: (a -> b -> c) -> b -> a -> c result in f (+) 1 2 evaluating to
by TheAsprngHacker 7y ago
Maybe I'm misunderstanding what you're saying, but how on Earth would a function
f :: (a -> b -> c) -> b -> a -> c
result in
f (+) 1 2
evaluating to
"cookie"
? For one, the types (obviously, to me at least) don't work out.
- wellpast 7y agoSounds like you could use some real documentation.