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I think that the misconception the author talks about happens because a noun is used for the properties of an object. An adjective such as "monadic" can be use

by spion 12y ago

I think that the misconception the author talks about happens because a noun is used for the properties of an object. An adjective such as "monadic" can be used instead: "In Haskell, IO actions are monadic because the IO type has a flatMap (bind) operator and a unit (return) function which satisfy left identity, right identity and associativity". All of the above are the properties (traits?) of the IO datatype. Monads (in Haskell) don't "exist" by themselves, they're just a property.

mattdw 12y ago
It took me so long to figure out that "where is the monad?" was the wrong question. It took me even longer to grok that "monad" is essentially an interface/trait rather than a "type". I think your suggestion of 'monadic' is a big improvement.
tel 12y ago
This is pretty true and certainly important but "Monad" does exist for many reasons. It's essentialy standing for "a monadic type" used when you're given a value of some monadic type but no specific one. This happens (in various forms) a lot in Haskell and so the term has weight. But for learning purposes I agree wholeheartedly!
dllthomas 12y ago
I think a part of it is a slight weirdness of Haskell typeclasses. In math, the noun "monad" refers to is a triple containing a functor and two natural transformations. The equivalent noun in Haskell would be the typeclass instance for a particular type. However, because Haskell (basically) only allows one instance per type, you select different instances by choosing a different (possibly equivalent, as with newtype) type, and so there is some natural conflation that arises in the language people use to talk about these things. The dynamic may be slightly clearer for semigroups. In mathematics, a semigroup is a pair containing a set and an (associative) operation on that set. In Haskell we say things like "ByteString is a semigroup" but that's short for "there is a semigroup where the set is all possible ByteStrings" and you assume the people you're talking to understand what the operation is because there is only one instance of Semigroup commonly defined for ByteString.

by spion 12y ago

mattdw 12y ago

It took me so long to figure out that "where is the monad?" was the wrong question. It took me even longer to grok that "monad" is essentially an interface/trait rather than a "type". I think your suggestion of 'monadic' is a big improvement.

tel 12y ago

This is pretty true and certainly important but "Monad" does exist for many reasons. It's essentialy standing for "a monadic type" used when you're given a value of some monadic type but no specific one. This happens (in various forms) a lot in Haskell and so the term has weight. But for learning purposes I agree wholeheartedly!

dllthomas 12y ago

I think a part of it is a slight weirdness of Haskell typeclasses. In math, the noun "monad" refers to is a triple containing a functor and two natural transformations. The equivalent noun in Haskell would be the typeclass instance for a particular type. However, because Haskell (basically) only allows one instance per type, you select different instances by choosing a different (possibly equivalent, as with newtype) type, and so there is some natural conflation that arises in the language people use to talk about these things. The dynamic may be slightly clearer for semigroups. In mathematics, a semigroup is a pair containing a set and an (associative) operation on that set. In Haskell we say things like "ByteString is a semigroup" but that's short for "there is a semigroup where the set is all possible ByteStrings" and you assume the people you're talking to understand what the operation is because there is only one instance of Semigroup commonly defined for ByteString.