Haskell for some reasons insists that I should only go for "The concept of a monad, which arises from category theory, has been applied by Moggi to structure the denotational semantics of programming languages" and
A monad is a triple (M,unit,⋆) consisting of a type constructor M and two operations of the given
polymorphic types. These operations must satisfy three laws given in Section 3.
We will often write expressions in the form
m ⋆ λa. n
Should I? Really?They allow you to edit code without the aforementioned released-last-week test runner having to check all your code after a big refactor. I mean, you can't call something with a "flatMap" method and a "return" method a monad! There are tons of nonsensical definitions that fit that which are going to become very unpleasant to use quickly.
Oh wow. How did I ever live with big refactorings before?
> There are tons of nonsensical definitions that fit that which are going to become very unpleasant to use quickly.
I recently learned that, apparently, I've been using "monads" and "monadic composition" for years now, never knowing what it is. Can't remember any unpleasantness that would "quickly arise".
There are other things than blind following after a Platonic ideal.
xs >>= f = concat (reverse (map f xs))
or Array.prototype.chain = function (f) {
return reverse(this.reduce((acc, it) => acc.concat(f(it)), []))
}
(I don't know how JS works, but you get it) instead of xs >>= f = concat (map f xs)
and enjoy refactoring your code, blissfully and consciously ignorant of the laws that make a monad a monad. The laws aren't just supposed to enrich the "life of the mind".Suppose you're going over some code, where you have something of the form
xs >>= someFunction
(where xs is a list) and you change someFunction to "return": xs >>= return
Then you'd expect, from experience, that you can rewrite this to just xs
(which you can check works with all the monads you know about) but that doesn't hold for this wonky definition of the Monad instance for lists. Indeed, note that return for lists has the definition return x = [x]
so, with our bad >>=, [1, 2, 3] >>= return
= concat (reverse (map (\x -> [x]) [1, 2, 3]))
= concat (reverse ([[1], [2], [3]]))
= concat [[3], [2], [1]]
= [3, 2, 1]
which is not the same as [1, 2, 3].