Haskell:Double和Int上的操作给出错误“无法匹配预期类型”

时间:2013-08-14 19:54:43

标签: haskell types instance

我正在为自己的语言编写一个解释器,我有一个抽象语法树,它有这种类型:

data Expression = 
  PInt Int
  | PFloat Double
  | PString String
  | PChar Char
  | PBool Bool
  | Var String
  | Unbound String String
  | Unary String Expression
  | Binary String Expression Expression
  | Call Expression [Expression]
  | Lambda Expression
  | Assign String Expression Expression
  | Conditional Expression Expression Expression
  deriving Eq

我正在尝试为我的班级编写一个Num实例,以便我可以使用现有的机器进行数值运算。这就是我写的:

instance Num Expression where
   PInt a + PInt b = PInt $ a + b
   PInt a + PFloat b = PFloat $ a + b
   PFloat a + PInt b = PFloat $ a + b
   PFloat a + PFloat b = PFloat $ a + b
   _ + _ = undefined
   PInt a - PInt b = PInt $ a - b
   PInt a - PFloat b = PFloat $ a - b
   PFloat a - PInt b = PFloat $ a - b
   PFloat a - PFloat b = PFloat $ a - b
   _ - _ = undefined
   PInt a * PInt b = PInt $ a * b
   PInt a * PFloat b = PFloat $ a * b
   PFloat a * PInt b = PFloat $ a * b
   PFloat a * PFloat b = PFloat $ a * b
   _ * _ = undefined
   negate (PInt a) = PInt (-a)
   negate (PFloat a) = PFloat (-a)
   negate _ = undefined
   abs (PInt a) = PInt $ abs a
   abs (PFloat a) = PFloat $ abs a
   abs _ = undefined
   signum (PInt a) = PInt $ signum a
   signum (PFloat a) = PFloat $ signum a
   signum _ = undefined
   fromInteger i = (PInt $ fromInteger i)

这给了我特别在我合并整数和浮点数的地方的错误。

Prelude> :load AST.hs
[1 of 1] Compiling AST          ( AST.hs, interpreted )

AST.hs:38:36:
    Couldn't match expected type `Double' with actual type `Int'
    In the first argument of `(+)', namely `a'
    In the first argument of `PFloat', namely `(a + b)'
    In the expression: PFloat (a + b)

AST.hs:39:37:
    Couldn't match expected type `Double' with actual type `Int'
    In the second argument of `(+)', namely `b'
    In the second argument of `($)', namely `a + b'
    In the expression: PFloat $ a + b

AST.hs:43:33:
    Couldn't match expected type `Double' with actual type `Int'
    In the first argument of `(-)', namely `a'
    In the second argument of `($)', namely `a - b'
    In the expression: PFloat $ a - b

AST.hs:44:37:
    Couldn't match expected type `Double' with actual type `Int'
    In the second argument of `(-)', namely `b'
    In the second argument of `($)', namely `a - b'
    In the expression: PFloat $ a - b

AST.hs:48:33:
    Couldn't match expected type `Double' with actual type `Int'
    In the first argument of `(*)', namely `a'
    In the second argument of `($)', namely `a * b'
    In the expression: PFloat $ a * b

AST.hs:49:37:
    Couldn't match expected type `Double' with actual type `Int'
    In the second argument of `(*)', namely `b'
    In the second argument of `($)', namely `a * b'
    In the expression: PFloat $ a * b
Failed, modules loaded: none.

这对我来说没有意义,因为Haskell中Int + Double的类型是Double,所以a + b应该解析为Double,因为PFloat的构造函数需要Double,没问题。 ..为什么不是这样?

已解决:在fromIntegral类型的变量前面使用Int修复它。

1 个答案:

答案 0 :(得分:3)

Num类型类中的数学运算符期望它们的两个参数具有相同的类型,因此您必须先使用fromIntegral将Int转换为Double,然后才能将它们添加到一起。

例如,替换此

PInt a + PFloat b = PFloat $ a + b

用这个

PInt a + PFloat b = PFloat $ fromIntegral a + b