archives

Prove: 'Cont r a = (a -> r) -> r' forms a monad

I don't follow Haskell too much, and more often than not I disagree with Erik.

However, this came up during an interview so it's supposedly something I should know.

Prove that 'Cont r a = (a -> r) -> r' forms a monad.

Any takers?

By marco at 2017-05-11 20:37 | LtU Forum | 47 comments | other blogs | 5920 reads