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Comment by jes5199 | Hacker News Reader

I know monads are supposed to be simple, but I've been playing with Haskell on and off for years now and I still can't figure out how do-notation desugars into >>= operations.

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It's not intuitive, but it's straightforward once you understand the semantics. There are basically two cases. Case 1:

    x <- monadicFunctionA
    monadicFunctionB

is rewritten as:

    monadicFunctionA >>= \x -> monadicFunctionB

The important thing here is that x is just the variable name in an lambda expression that you don't see. Case 2:

    monadicFunctionA
    monadicFunctionB

is rewritten as:

    monadicFunctionA >> monadicFunctionB

Where >> === >>= \_ ... . Given these rules, you can approximate do-notation like so:

    monadicFunctionA >>=
    \x -> monadicFunctionB >>
    monadicFunctionC ...

    do
      a
      rest

desugars to

    a >>
    do
      rest

and

    do
      x <- a
      rest

desugars to

    a >>= \x ->
    do
      rest

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