Re: [Haskell-cafe] Monad explanation

Huh? You can't actually over-ride function application --- Haskell's built-in application always does exactly the same thing, at every type.

It's a metaphor. In every case container has its own method of applying functions to its contents - e.g. instead of "f x" we write "x fmap f".

You can, however, define new application-like operators that have other application-like behaviors. Haskell has a few of these:

* return * (<$>) * (<*>) * (>>=)

I know Haskell, that's not what I'm talking about. What I'm trying to do is to create a simple explanation for monads that doesn't involve neither mathematics nor "computation" metaphor.