monads

问题 I'm trying to declare MonadPlus interface like that: module NanoParsec.Plus %access public export interface Monad m => MonadPlus m where zero : m a plus : m a -> m a -> m a But have an error: | 5 | interface Monad m => MonadPlus m where | ~~~~~~~ When checking type of constructor of NanoParsec.Plus.MonadPlus#Monad m: When checking argument m to type constructor Prelude.Monad.Monad: Type mismatch between Type (Type of m) and Type -> Type (Expected type) What I'm doing wrong? How to fix this?

问题 In this question the OP asks what the type of the expression return 5 is and the answer has already been given in that question: it is a generic type, as can be verified by typing :t return 5 in the Haskell interpreter: return 5 :: (Num a, Monad m) => m a The specific implementation of return is determined by the context in which it appears: type inference will restrict m to a specific monad such as Maybe , [] , IO , and so on. I can also force the interpreter to pick a specific monad by

问题 In this question the OP asks what the type of the expression return 5 is and the answer has already been given in that question: it is a generic type, as can be verified by typing :t return 5 in the Haskell interpreter: return 5 :: (Num a, Monad m) => m a The specific implementation of return is determined by the context in which it appears: type inference will restrict m to a specific monad such as Maybe , [] , IO , and so on. I can also force the interpreter to pick a specific monad by

问题 Rust does not have higher-kinded-types. For example, functor (and thus monad) cannot be written in Rust. I would like to know if there is a deep reason explaining this and why. For instance, reason that I can understand can be that there is no zero-cost abstraction making HKT possible. Or type inference is significantly more difficult. And of course, I also looking for an explaination showing me why this is a real limitation. If the anwer was already given somewhere else, could you give me

问题 Rust does not have higher-kinded-types. For example, functor (and thus monad) cannot be written in Rust. I would like to know if there is a deep reason explaining this and why. For instance, reason that I can understand can be that there is no zero-cost abstraction making HKT possible. Or type inference is significantly more difficult. And of course, I also looking for an explaination showing me why this is a real limitation. If the anwer was already given somewhere else, could you give me

问题 Rust does not have higher-kinded-types. For example, functor (and thus monad) cannot be written in Rust. I would like to know if there is a deep reason explaining this and why. For instance, reason that I can understand can be that there is no zero-cost abstraction making HKT possible. Or type inference is significantly more difficult. And of course, I also looking for an explaination showing me why this is a real limitation. If the anwer was already given somewhere else, could you give me

问题 Thinking about monad, it came to me the idea of a monad as the way to break with the Von Neumann architecture. The Von Neumann architecture uses a set of instructions (called program) to change the data in memory and the execution of each instruction of the program updates a program counter to know whom instruction is the next to execute. If we think about the Von Neumann architecture as a monad, the bind operator (>>=) update the program counter. We can make a Monad that break Von Neumann

问题 as I had been looking for ways to optimize a password cracker I had been making, I came across a much shorter implementation of all possible character combinations in a list, which used this function: mapM (const xs) [1..n] where xs could be the characters available, and n the length of the desired words. so mapM (const "abcd") [1..4] would output a list ["aaaa","aaab","aaac","aaad","aaba","aabb"..] and so on. Only the length matters for the list of th right, I could have written ['f','h','s'

来源： https://stackoverflow.com/questions/59869399/why-does-mutual-yielding-make-arrowapply-and-monads-equivalent-unlike-arrow-and

来源： https://stackoverflow.com/questions/59869399/why-does-mutual-yielding-make-arrowapply-and-monads-equivalent-unlike-arrow-and