3 ms·

Yes, you should care about the laws! They allow you to edit code without the aforementioned released-last-week test runner having to check all your code after

Yes, you should care about the laws! 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.

dmitriid 9y ago
> They allow you to edit code without the aforementioned released-last-week test runner having to check all your code after a big refactor. 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.
- mrkgnao 9y ago
  The unpleasantness arises when writing Monad instances that don't follow the laws. Setting your disingenuity aside, try using 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".
  - dmitriid 9y ago
    So, what problems will I experience while refactoring this code (if I ever wrote it)?
    - mrkgnao 9y ago
      It won't satisfy the properties you've come to expect from the implementations you know for lists, optionals (Maybe), and so on. 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]. https://wiki.haskell.org/Monad_laws https://wiki.haskell.org/Monad_laws

dmitriid 9y ago

> They allow you to edit code without the aforementioned released-last-week test runner having to check all your code after a big refactor. 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.

mrkgnao 9y ago
The unpleasantness arises when writing Monad instances that don't follow the laws. Setting your disingenuity aside, try using 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".
- dmitriid 9y ago
  So, what problems will I experience while refactoring this code (if I ever wrote it)?
  - mrkgnao 9y ago
    It won't satisfy the properties you've come to expect from the implementations you know for lists, optionals (Maybe), and so on. 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]. https://wiki.haskell.org/Monad_laws https://wiki.haskell.org/Monad_laws