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No, but Haskell does not need to explain monad for that either. All examples and docs for map/concatMap over a list are free from having to mention any type cla
by jose_zap 4y ago
No, but Haskell does not need to explain monad for that either. All examples and docs for map/concatMap over a list are free from having to mention any type class at all.
On the other hand, if you have to explain how the for/yield in zio and why it works for so many different types, you have to basically explain Monad using another name.
- valenterry 4y ago> On the other hand, if you have to explain how the for/yield in zio and why it works for so many different types, you have to basically explain Monad using another name. But again, that's not happening.