Pass Types as arguments to a function in Haskell?

haskell, types

Solution

Haskell supports a high degree of polymorphism. In particular

readAny n stream = foldl next ([], stream) [1..n]
  where
    next (lst, x:xs) _ = (lst ++ [v], xs)
      where
        v = read x

has type

readAny :: (Enum b, Num b, Read a) => b -> [String] -> ([a], [String])

thus

readInts :: (Enum b, Num b) => b -> [String] -> ([Int], [String])
readInts = readAny

readFloats :: (Enum b, Num b) => b -> [String] -> ([Float], [String])
readFloats = readAny

you dont need to specialize the type. Haskell will automatically infer the most general type possible, and the `readAny` here will do what you want.

It is not possible to pass types as arguments in Haskell. Rarely would you need to. For those few cases where it is necessary you can simulate the behavior by passing a value with the desired type.

Haskell has "return type polymorphism" so you really shouldn't worry about "passing the type"--odds are that functions will do what you want without you telling them to.

Problem

The following two functions are extremely similar. They read from a [String] n elements, either [Int] or [Float]. How can I factor the common code out? I don't know of any mechanism in Haskell that supports passing types as arguments. ``` readInts n stream = foldl next ([], stream) [1..n] where next (lst, x:xs) _ = (lst ++ [v], xs) where v = read x :: Int readFloats n stream = foldl next ([], stream) [1..n] where next (lst, x:xs) _ = (lst ++ [v], xs) where v = read x :: Float ``` I am at a beginner level of Haskell, so any comments on my code are welcome.

Original source