Haskells 绑定算子和>>算子及其关系

问：

我最近发布了一个关于操作员的问题，因为即使我已经阅读了 LYAH walk the linee 部分，我的理解仍然存在一些差距。以下是我偶然发现的一些代码/MVE，从那时起，它引发了以下思考。为什么我可以得到代码后面的输出？难道不是没有向 Bind 运算符提供任何参数，因此没有参数可以连接，否则它们将使用 bind，如定义所示，并且其结果不应与下面提到的预期结果相似：>>str>>=

import Control.Monad
import Data.List

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

data RErr = EBVar String  | EBInt Int
  deriving (Eq, Show)

newtype Compuptaton a = Computation {runComputation :: [(String, Value)] -> (Either RErr a, [String]) }

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

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

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

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

输出：

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

预期输出

(Right NoneVal, [])

此外，为什么会用额外的串联来扩展绑定运算符定义：str'

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

而另一个像这样： ，不会导致以下结果：apply>>apply "print" values = output (unwords [showValue x| x<-values]) >> output (unwords [showValue x| x<-values]) >> return NoneVal

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

而不是实际的：

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

只有 3 个输入元素。