Pass Types as arguments to a function in Haskell?
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.
Basically what you want is to not explicitly declare the type. Instead, defer declaring the type and let the inference engine take over for you. Also, I think you are conflating fold with map. This is how I would approach it.
readList' :: Read a => [String] -> [a]
readList' = map read
ints = readList' ["1", "2"] :: [Int] -- [1, 2]
floats = readList' ["1.0", "2.0"] :: [Float] -- [1.0, 2.0]
To read only n things from the stream, use take