newtype: Can someone explain me, how this code works?

Question

I defined new type like "PositiveInteger" like below.

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

When i execute the statement 'fromPositiveInteger (10)', i am getting following error. Which is absolutely fine.

*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)

Suppose, if i updated my code like below. the statement 'fromPositiveInteger 10' works without any error. How can this happened?

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，因此10 （其最通用的类​​型是Num a => a ）可以具有类型PositiveInteger ，使其成为fromPositiveInteger的有效参数。