我最近发布了一个关于>>运算符的问题，因为即使我已经阅读了LYAH walk the linee部分，我仍然有一些理解上的差距。下面是我偶然发现的一些代码/MVE，它们引发了以下思考。为什么我可以得到代码后面的输出？如果bind操作符没有提供参数，那么str参数就不能连接起来，否则它们将使用bind，如>>=定义中所示，结果不应该与下面提到的预期结果相似：

import Control.Monad
import Data.List

data Value =
    NoneVal
  | IntVal Int
  | ListVal [Value]
  deriving (Eq, Show, Read)

type FName = String

data RError = EBadVar String | EBadFun String | EBadArg String
  deriving (Eq, Show)

newtype Comp a = Comp {runComp :: [(String, Value)] -> (Either RError a, [String]) }

instance Monad Comp where
  return a = Comp( \_ -> (Right a, []))
  m >>= f = Comp(\env -> case runComp m env of
        (Left e, str) -> (Left e, str)
        (Right a, str) -> let (a', str') = runComp (f a) env in (a', str++str'))

-- You shouldn't need to modify these
instance Functor Comp where
  fmap = liftM
instance Applicative Comp where
  pure = return; (<*>) = ap

showValue :: Value -> String
showValue (IntVal int) = show int
showValue (ListVal values) = "[" ++ intercalate ", " [showValue x| x<-values] ++ "]"

output :: String -> Comp ()
output s = Comp (\_ -> (Right (), [s]))

-- Helper functions for interpreter

apply :: FName -> [Value] -> Comp Value
apply "print" values = output (unwords [showValue x| x<-values]) >> return NoneVal

字符串
输出量：

ghci> runComp (apply "print" [(IntVal 1), (IntVal 2)]) [("x", (IntVal 4)), ("y", (IntVal 3))]
(Right NoneVal,["1 2"])

型
预期产量

(Right NoneVal, [])

型
此外，为什么要用一个额外的str'串联来扩展bind运算符定义：

(Left e, str) -> (Left e, str)
        (Right a, str) -> let (a', str') = runComp (f a) env in (a', str++str'++str'))

型
apply和另一个>>，如下所示：apply "print" values = output (unwords [showValue x| x<-values]) >> output (unwords [showValue x| x<-values]) >> return NoneVal，不会导致以下结果：

ghci> runComp (apply "print" [(IntVal 1), (IntVal 2)]) [("x", (IntVal 4)), ("y", (IntVal 3))]
(Right NoneVal,["1 2","1 2","1 2", "1 2"])

型
而不是实际的：

ghci> runComp (apply "print" [(IntVal 1), (IntVal 2)]) [("x", (IntVal 4)), ("y", (IntVal 3))]
(Right NoneVal,["1 2","1 2","1 2"])

型
只有3个输入元素。