Re: [Haskell-cafe] mtl: Why there is "Monoid w" constraint in the definition of class MonadWriter?

The only thing we can tell from the Monad laws is that that function f should be associative.

That f is associative is a very small step away from f forming a monoid. What about listen :: m a -> m (w, a)? What laws should it hold that are compatible with those of the monad and those of tell? Reasoning about listen is enough to force the existence of a monoid identity. -- Kim-Ee On Sun, Dec 9, 2012 at 5:41 AM, Roman Cheplyaka wrote:

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Excerpts from Roman Cheplyaka's message of Sat Dec 08 14:00:52 -0800 2012:

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* Edward Z. Yang [2012-12-08 11:19:01-0800]

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The monoid instance is necessary to ensure adherence to the monad laws.

This doesn't make any sense to me. Are you sure you're talking about

* Edward Z. Yang [2012-12-08 14:18:38-0800] the

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MonadWriter class and not about the Writer monad?

Well, I assume the rules for Writer generalize for MonadWriter, no?

Here's an example. Haskell monads have the associativity law:

(f >=> g) >=> h === f >=> (g >=> h)

From this, we can see that

(m1 >> m2) >> m3 === m1 >> (m2 >> m3)

Now, consider tell. We'd expect it to obey a law like this:

tell w1 >> tell w2 === tell (w1 <> w2)

First of all, I don't see why two tells should be equivalent to one tell. Imagine a MonadWriter that additionally records the number of times 'tell' has been called. (You might argue that your last equation should be a MonadWriter class law, but that's a different story — we're talking about the Monad laws here.)

Second, even *if* the above holds (two tells are equivalent to one tell), then there is *some* function f such that

tell w1 >> tell w2 == tell (f w1 w2)

It isn't necessary that f coincides with mappend, or even that the type w is declared as a Monoid at all. The only thing we can tell from the Monad laws is that that function f should be associative.

Roman

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