Anyone know the answer to this? I was just reminded in another thread that this order prevents the definition of the (>>) operator, which sequences 2 monadic computations. Instead, we have to use the ugly >>= fun () -> paradigm.
The reason we can’t have (>>) is just that the right hand side (the latter part of the computation) would be evaluated before the left hand side (the former part of the computation). Of course, we can have (<<), but given the way monadic computation is constructed, I don’t think it’s very useful.