Haskell从do符号到Applicative符号的传递(第2部分)

我试图去掉数据库中的do符号。sh文件来自 https://haskell-at-work.com/episodes/2018-01-19-domain-modelling-with-haskell-data-structures.html

但我有一个错误,我不知道为什么。(可能只是意味着我不知道哈斯克尔的事)

这是本书的延续 Haskell Passing from do notation to Applicative

哈斯克尔代码:

项目hs

{-# LANGUAGE GeneralizedNewtypeDeriving #-}
module Project where

import           Data.Text (Text)

newtype Money = Money
  { unMoney :: Double
  } deriving (Show, Eq, Num)

newtype ProjectId = ProjectId
  { unProjectId :: Int
  } deriving (Show, Eq, Num)

data Project
  = Project ProjectId
            Text
  | ProjectGroup Text
                 [Project]
  deriving (Show, Eq)

data Budget = Budget
  { budgetIncome      :: Money
  , budgetExpenditure :: Money
  } deriving (Show, Eq)

data Transaction
  = Sale Money
  | Purchase Money
  deriving (Eq, Show)

数据库

import           System.Random (getStdRandom, randomR)

import Project

getBudget :: ProjectId -> IO Budget
getBudget _ = Budget
    <$> (Money <$> getStdRandom (randomR (0, 10000)))
    <*> (Money <$> getStdRandom (randomR (0, 10000)))

getTransactions :: ProjectId -> IO [Transaction]
getTransactions _ = 
  let rtn = []
       <$>  (Sale . Money <$> getStdRandom (randomR (0, 4000)))
       <*> (Purchase . Money <$> getStdRandom (randomR (0, 4000)))
  in
    pure rtn

错误

跑步后 stack ghc Database.hs --package random

Database.hs:12:13: error:
    * Couldn't match expected type: Transaction -> Transaction -> b
                  with actual type: [a0]
    * In the first argument of `(<$>)', namely `[]'
      In the first argument of `(<*>)', namely
        `[] <$> (Sale . Money <$> getStdRandom (randomR (0, 4000)))'
      In the expression:
        [] <$> (Sale . Money <$> getStdRandom (randomR (0, 4000)))
          <*> (Purchase . Money <$> getStdRandom (randomR (0, 4000)))
    * Relevant bindings include rtn :: f b (bound at Database.hs:12:7)
   |
12 |   let rtn = []
   |             ^^