在 Haskell 中将 25 美分、10 美分和便士转换为美元和美分

Question

我正在尝试将 25 美分、10 美分和便士转换为美元和美分，但我似乎无法运行代码。

我试过这段代码：

-- Declaration
toCents :: Int -> Int -> Int -> Int
toDollar :: Int -> Int -> Int -> Int

-- Definition
toCents quarter dime penny  = quarter*25 + dime*10 + penny
toDollar quarter dime penny = (toCents quarter dime penny) / 100) - ((toCents quarter dime penny mod 100) / 100)

main :: IO()
main = do
 let quarter = 12
 let dime = 67
 let penny = 43
 let sumDollar = toDollar quarter dime penny
 let sumCents = toCents quarter dime penny - (toDollar quarter dime penny) * 100
 print ("Conversion of " ++ show quarter ++ " quarters, " ++ show dime ++ " dimes, and " ++ show penny ++ " pennies to dollars and cents:")
 print(sumDollar ++ " dollars and " ++ sumCents ++ " cents.")

它导致了这个错误：

[1 of 1] Compiling Main             ( main.hs, main.o )
main.hs:7:68: error:
    • Couldn't match expected type ‘(Integer -> Integer -> Integer)
                                    -> Integer -> Int’
                  with actual type ‘Int’
    • The function ‘toCents’ is applied to five arguments,
      but its type ‘Int -> Int -> Int -> Int’ has only three
      In the first argument of ‘(/)’, namely
        ‘(toCents quarter dime penny mod 100)’
      In the second argument of ‘(-)’, namely
        ‘((toCents quarter dime penny mod 100) / 100)’
  |
7 | toDollar quarter dime penny = (toCents quarter dime penny / 100) - ((toCents quarter dime penny mod 100) / 100)
  |                                                                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
main.hs:17:8: error:
    • Couldn't match expected type ‘[Char]’ with actual type ‘Int’
    • In the first argument of ‘(++)’, namely ‘sumDollar’
      In the first argument of ‘print’, namely
        ‘(sumDollar ++ " dollars and " ++ sumCents ++ " cents.")’
      In a stmt of a 'do' block:
        print (sumDollar ++ " dollars and " ++ sumCents ++ " cents.")
   |
17 |  print(sumDollar ++ " dollars and " ++ sumCents ++ " cents.")
   |        ^^^^^^^^^
main.hs:17:40: error:
    • Couldn't match expected type ‘[Char]’ with actual type ‘Int’
    • In the first argument of ‘(++)’, namely ‘sumCents’
      In the second argument of ‘(++)’, namely ‘sumCents ++ " cents."’
      In the second argument of ‘(++)’, namely
        ‘" dollars and " ++ sumCents ++ " cents."’
   |
17 |  print(sumDollar ++ " dollars and " ++ sumCents ++ " cents.")
   |                   

Answer 1

表达式toCents quarter dime penny mod 100被解析为

     ((((toCents quarter) dime) penny) mod) 100

或者

      ( ( ( (toCents quarter
            ) dime
          ) penny
        ) mod
      ) 100

这看起来很奇怪，但如果你记住柯里化是如何工作的，它并没有那么奇怪：它对应于，在一种非柯里化语言中，

      toCents(quarter, dime, penny, mod, 100)

这当然仍然是无稽之谈：您将mod作为参数提供给 function toCents ，而不是将其用作运算符。 带有字母名称的中缀运算符必须写在反引号中才能应用它们，例如37`mod`4 ，与mod 37 4相同。 中缀运算符总是绑定弱于前缀 function 应用程序，所以如果你写了

  toCents quarter dime penny `mod` 100

那么这将被解析为

  mod (toCents quarter dime penny) 100

这可能是你想要的。 然而，这并不意味着它适合这项任务。 你在做什么

    foo/100 - (foo`mod`100)/100

是实现简单舍入到负数 integer 除法的非常迂回的尝试。

它在 Python 中有点工作，但即使在那里它也没有真正做它应该做的事情：Python 自动将整数foo和100转换为浮点数以执行除法，然后相互减去浮点数，结果仍然是浮点数它恰好有零小数部分。

Haskell 甚至不允许这种混乱发生：它会抱怨/运算符在整数上不可用。 你真的想要那个，你必须明确地转换为浮点数。

但是您不希望这样：您应该简单地使用内置的 integer 除法运算符

toDollar quarter dime penny = toCents quarter dime penny `div` 100

...假设它真的应该有舍入到消极的行为。

更好的方法是正确构建代码和数据，这将避免冗余计算和神秘的Int -> Int ->...签名：

data DollarValue = DollarValue {
       dollarPartVal, centPartVal :: Int
     }
data DollarCoins = DollarCoins {
       quarterCoinCount, dimeCoinCount, pennyCoinCount :: Int
     }

coinsValue :: DollarCoins -> DollarValue
coinsValue (DollarCoins q d p) = DollarValue dollars cents
 where (dollars, cents) = (25*q + 10*d + p)`divMod`100

import Text.Printf

main = do
   let quarter = 12
       dime = 67
       penny = 43
   let DollarValue sumDollars sumCents
        = coinsValue $ DollarCoins quarter dime penny
   printf
    "Conversion of %i quarters, %i dimes, and %i pennies: %i dollars and %i cents.\n"
                   quarter      dime          penny       sumDollars     sumCents