newtype：有人可以解释一下，这段代码是如何工作的？

Question

我定义了类似“PositiveInteger”的新类型，如下所示。

newtype PositiveInteger = PositiveInteger Integer deriving Show

fromPositiveInteger :: PositiveInteger -> Integer
fromPositiveInteger (PositiveInteger i) = i

toPositiveInteger :: Integer -> PositiveInteger
toPositiveInteger x
    | (x < 0) = error "Not applicable to negative numbers"
    | otherwise = PositiveInteger x

当我执行语句'fromPositiveInteger（10）'时，我收到以下错误。这绝对没问题。

*Main> fromPositiveInteger (10)

<interactive>:7:22:
    No instance for (Num PositiveInteger) arising from the literal ‘10’
    In the first argument of ‘fromPositiveInteger’, namely ‘(10)’
    In the expression: fromPositiveInteger (10)
    In an equation for ‘it’: it = fromPositiveInteger (10)

假设，如果我更新了我的代码，如下所示。 'fromPositiveInteger 10'语句没有任何错误。怎么会发生这种情况？

newtype PositiveInteger = PositiveInteger Integer deriving Show

fromPositiveInteger :: PositiveInteger -> Integer
fromPositiveInteger (PositiveInteger i) = i

toPositiveInteger :: Integer -> PositiveInteger
toPositiveInteger x
    | (x < 0) = error "Not applicable to negative numbers"
    | otherwise = PositiveInteger x

instance Num PositiveInteger where
    fromInteger         = toPositiveInteger
    x + y               = toPositiveInteger (fromPositiveInteger x + fromPositiveInteger y)
    x - y               = let r = fromPositiveInteger x - fromPositiveInteger y in
                            if r < 0 then error "Unnatural subtraction"
                                     else toPositiveInteger r
    x * y               = toPositiveInteger (fromPositiveInteger x * fromPositiveInteger y)

*Main> fromPositiveInteger (PositiveInteger 10)
10
*Main> fromPositiveInteger 10
10

Answer 1

由于PositiveInteger现在是Num的实例，10（其最常见类型为Num a => a）可以使用PositiveInteger类型，因此fromPositiveInteger的有效参数。