Hacker News Reader

Comment by pdpi | Hacker News Reader

Nah, it's not really "one of those famous misconceptions".

The crucial detail is that TardisT must wrap a MonadFix, which is used for cyclical data structures. It effectively works not by making time go "backwards", but by making time loop around. It still creates a sequencing relationship.

View on HN

> The crucial detail is that TardisT must wrap a MonadFix

So? The non-transformer version can still "read data before it's written". It's very hard to see any form of sequencing in that example.

> It still creates a sequencing relationship.

A sequencing relationship between what exactly?

The sequencing is described precisely in the signature of bind.

    (>>=)
        :: Monad m
        => m a         -- Left argument
        -> (a -> m b)  -- Right argument
        -> m b

You don’t need to explain an individual monad instance to notice that this signature requires “a” in order to generate “m b”.

What about the Identity monad? Does that "force sequencing"? What "sequencing relationship" does it create?

If you are convinced that the Identity monad does "force sequencing" then you are also convinced that lazy evaluation "forces sequencing" (because the Identity monad is just lazy evaluation). That would be a fairly controversion conviction. If you are not convinced that the Identity monad "forces sequencing" then can you explain in more detail the meaning of your claim "What a Monad does is to force sequencing by creating a data dependency.".

The identity monad's sequencing is just basic cause and effect: you need to have produced an `a` in order to get a `b`.

The sequencing is very obvious from the signature of `>>=`. You don't need to know anything about any instances to see that it's there.

> The identity monad's sequencing is just basic cause and effect: you need to have produced an `a` in order to get a `b`.

You most certainly do not! See https://news.ycombinator.com/item?id=16420742

It's in the signature. `b` has a causal dependency on `a` because `a` is on the left hand side of the arrow and `b` is on the right hand side of the arrow. That's it. Functional programming 101. There's no way around this.

The operational semantics and details of thunks are not relevant here. Tricks with `undefined` are not relevant here. This is just the meaning of the arrow type.

If that were true, why do you need a monad for sequencing? Why doesn’t function composition suffice, or something of the form a -> (a -> b) -> b?

Because the values have extra "stuff" surrounding them. That's why it's `m a` and not just `a`: the `m` holds the invisible context that enriches the `a`.

Although your example is exactly what's happening in the identity monad, because there is no extra stuff.

Monads have been called "executable semicolons", because you can define a function to be executed each time the program passes from one statement to the next.

That's not quite right.

Let's define an operator `|> = (flip $)`, which has a type signature `a -> (a -> b) -> b`, which, by your reasoning, should force sequencing too. Except it doesn't: Laziness is a bitch to work with, and `a |> f |> g` evaluates to `g(f(a))`, and g is evaluated first, which then evaluates f(a) on demand. If you consider `1 |> (+1) |> const 4`, this will evaluate to 4 without ever performing the addition.

Instead, let's compare this to the monadic equivalent. `(return 1) >>= return . (+ 1) >>= return . (const 4)`. The sequence that's being forced here isn't that (+ 1) is evaluated, it's that the `return` in `return 1` is evaluated (producing a 1) before `return . (+ 1)` is ever touched, and that _that_ return is evaluated (presumably producing an unevaluated thunk for (1 + 1)) before `return . (const 4)` is touched.

In IO world, this corresponds to guaranteeing that the IO operations happen in sequence irrespective of the order in which the pure computations they wrap are evaluated in.

> the `return` in `return 1` is evaluated (producing a 1) before `return . (+ 1)` is ever touched, and that _that_ return is evaluated (presumably producing an unevaluated thunk for (1 + 1)) before `return . (const 4)` is touched.

That's not quite right either. The former returns need never be evaluated.

    > undefined >>= (return . (const 4)) :: Identity Int
    Identity 4

Right — I should've said "that's the mechanism through which the implementation of `>>=` can force sequencing". `>>=` for Identity doesn't, most of the other common ones do, in some sense or another.

Right! `>>=` can force sequencing. It doesn't for Identity, Writer.Lazy or State.Lazy, but it can. OK, but also Arrow and Monoid can enforce a (different sort of) sequencing relationship.

Neither does monad force a sequencing relationship nor is it the only thing than can force a sequencing relationship so I can only conclude that "What a Monad does is to force sequencing by creating a data dependency" is indeed one of those misconceptions!

That's just Haskell's crazy evaluation semantics though, not something that applies to monads in general.

Indeed. But bad_user made a claim about monads in general ("What a Monad does is to force sequencing by creating a data dependency.") so all I need to do to refute it is give an example of a particular monad where the claim doesn't hold.

Monads force a denotation of sequencing. Of course you can write a language implementation where the sequencing of operation does not represent the sequencing of denotation (just as you can write a "compiler" that compiles all programs to Hello World), but, well, that's on you.

> Monads force a denotation of sequencing

Aha! Now we're on to something.

How is sequencing described in the type of join?

Tardis and the ReverseState monad are more examples of people being funny and clever by creating misleading names and mental-models instead of making libraries easy to understand.

There's no "time loop" in a cyclical data structure. There's just a data loop. Moving backward through a circular buffer, or doing arithmetic mod N, isn't "time travel"

Reply on news.ycombinator.com