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Rikitikitavi
Asked:2022-03-26 23:41:50 +0800 CST2022-03-26 23:41:50 +0800 CST

haskell 中的函数

  • 772

有这段代码，它试图描述一个机器人对象：

robot (name, attack, hp) = \message -> message(name, attack, hp)

name (n, _, _) = n
attack (_, a, _) = a
hp (_, _, hp) = hp

getAttack aRobot = aRobot attack
getHP aRobot = aRobot hp
getName aRobot = aRobot name

setName aRobot newName = aRobot(\(n, a, h) -> robot(newName, a , h))
setAttack aRobot newAttack = aRobot(\(n, a, h) -> robot(n, newAttack , h))
setHP aRobot newHP = aRobot(\(n, a, h) -> robot(n, n , newHP))

printRobot aRobot = aRobot( \(n, a, h) -> n ++ " attack: " ++ (show a) ++ " hp: " ++ (show h) )

damage aRobot dmg = aRobot(\(n,a,h) -> robot(n,a, h - dmg))

fight aRobot defender = damage defender dmg
    where dmg = if getHP aRobot > 0
                then getAttack aRobot
                else 0

getAllHP aRobots = map getHP aRobots

现在我需要编写一个函数，在一轮后返回两个机器人的元组，我这样做：

oneRoundFight aR1 aR2 = ( fight aR1 aR2, fight aR2 aR1 )

一个错误：

*Main> :l robot.hs
[1 of 1] Compiling Main             ( robot.hs, interpreted )

robot.hs:27:48: error:
    * Occurs check: cannot construct the infinite type:
        b2 ~ ((a2, b2, b1) -> t1) -> t1
      Expected type: ((a2, ((a2, b2, b1) -> t1) -> t1,
                       ((a2, b2, b1) -> t1) -> t1)
                      -> ((a2, b2, b1) -> t1) -> t1)
                     -> a1
        Actual type: ((a2, b2, b1) -> ((a2, b2, b1) -> t1) -> t1) -> a1
    * In the first argument of `fight', namely `aR2'
      In the expression: fight aR2 aR1
      In the expression: (fight aR1 aR2, fight aR2 aR1)
    * Relevant bindings include
        aR2 :: ((a2, b2, b1) -> ((a2, b2, b1) -> t1) -> t1) -> a1
          (bound at robot.hs:27:19)
        aR1 :: ((a, ((a, b, a1) -> t) -> t, ((a, b, a1) -> t) -> t)
                -> ((a, b, a1) -> t) -> t)
               -> b1
          (bound at robot.hs:27:15)
        oneRoundFight :: (((a, ((a, b, a1) -> t) -> t,
                            ((a, b, a1) -> t) -> t)
                           -> ((a, b, a1) -> t) -> t)
                          -> b1)
                         -> (((a2, b2, b1) -> ((a2, b2, b1) -> t1) -> t1) -> a1)
 -> (a1, b1)
          (bound at robot.hs:27:1)
   |
27 | oneRoundFight aR1 aR2 = ( fight aR1 aR2, fight aR2 aR1 )
   |                                                ^^^

robot.hs:27:52: error:
    * Occurs check: cannot construct the infinite type:
        b ~ ((a, b, a1) -> t) -> t
      Expected type: ((a, b, a1) -> ((a, b, a1) -> t) -> t) -> b1
        Actual type: ((a, ((a, b, a1) -> t) -> t, ((a, b, a1) -> t) -> t)
                      -> ((a, b, a1) -> t) -> t)
                     -> b1
    * In the second argument of `fight', namely `aR1'
      In the expression: fight aR2 aR1
      In the expression: (fight aR1 aR2, fight aR2 aR1)
    * Relevant bindings include
        aR2 :: ((a2, b2, b1) -> ((a2, b2, b1) -> t1) -> t1) -> a1
          (bound at robot.hs:27:19)
        aR1 :: ((a, ((a, b, a1) -> t) -> t, ((a, b, a1) -> t) -> t)
                -> ((a, b, a1) -> t) -> t)
               -> b1
          (bound at robot.hs:27:15)
        oneRoundFight :: (((a, ((a, b, a1) -> t) -> t,
                            ((a, b, a1) -> t) -> t)
                           -> ((a, b, a1) -> t) -> t)
                          -> b1)
                         -> (((a2, b2, b1) -> ((a2, b2, b1) -> t1) -> t1) -> a1)
 -> (a1, b1)
          (bound at robot.hs:27:1)
   |
27 | oneRoundFight aR1 aR2 = ( fight aR1 aR2, fight aR2 aR1 )
   |                                                    ^^^
Failed, no modules loaded.
Prelude>

我不太明白这个输出是什么意思，特别是cannot construct the infinite type: ，如果你改变地方的参数：( fight aR1 aR2, fight aR1 aR2 )那么这个错误就不会发生。

你能解释一下吗？

同时，函数fight编写并工作：

Prelude> :l robot.hs
[1 of 1] Compiling Main             ( robot.hs, interpreted )
Ok, one module loaded.
*Main> aR1 = robot("r1", 8, 100)
*Main> aR2 = robot("r2", 4, 120)
*Main> aR2_damaged = fight aR1 aR2
*Main> aR1_damaged = fight aR2 aR1
*Main> printRobot aR1_damaged
"r1 attack: 8 hp: 96"
*Main> printRobot aR2_damaged
"r2 attack: 4 hp: 112"
haskell

1 个回答

  1. Best Answer

    你确定吗。那为了存储这么简单的数据集，你真的需要这么复杂的数据类型（从函数到元组的函数）吗？

    如果是这样，那么您将不得不处理更高级别的多态性，而这几乎不是您现在想要处理的。

    如果没有，那么一切都可以更容易地描述。

    data Robot = Robot { robotName :: String, robotAttack :: Int, robotHP :: Int } deriving Show

getAttack = robotAttack
getName = robotName
getHP = robotHP

setAttack aRobot newAttack = aRobot { robotAttack = newAttack }
setHP aRobot newHP = aRobot { robotHP = newHP }
setName aRobot newName = aRobot { robotName = newName }

printRobot aRobot =
  getName aRobot ++
  " attack: " ++ show (getAttack aRobot) ++
  " hp: " ++ show (getHP aRobot)

damage aRobot dmg = setHP aRobot (getHP aRobot - dmg)

fight aRobot defender = damage defender dmg
    where dmg = if getHP aRobot > 0
                then getAttack aRobot
                else 0

getAllHP aRobots = map getHP aRobots

oneRoundFight aR1 aR2 = (fight aR2 aR1, fight aR1 aR2)

    GHCi> aR1 = Robot "r1" 8 100
GHCi> aR2 = Robot "r2" 4 120
GHCi> (aR1', aR2') = oneRoundFight aR1 aR2
GHCi> printRobot aR1'
"r1 attack: 8 hp: 96"
GHCi> printRobot aR2'
"r2 attack: 4 hp: 112"

    “没有额外的实体”

    oneRoundFight aR1 aR2 =
  aR1 $ \(n1, a1, h1) ->
    aR2 $ \(n2, a2, h2) ->
      (robot (n1, a1, h1 - dmg h2 a2), robot (n2, a2, h2 - dmg h1 a1))
  where dmg hp attack = if hp > 0 then attack else 0

    GHCi> aR1 = robot ("r1", 8, 100)
GHCi> aR2 = robot ("r2", 4, 120)
GHCi> (aR1', aR2') = oneRoundFight aR1 aR2
GHCi> printRobot aR1'
"r1 attack: 8 hp: 96"
GHCi> printRobot aR2'
"r2 attack: 4 hp: 112"
    • 3

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