How does `flip id` work?

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0

Why is flip id accepted as valid expression? Shouldn't it fail at type check of a against b -> c?

Prelude> :t flip
flip :: (a -> b -> c) -> b -> a -> c
Prelude> :t id
id :: a -> a
Prelude> :t flip id
flip id :: b -> (b -> c) -> c

asked 2 mins ago

sevo

459

How does `flip id` work?