Question

I have tried

map show . mapMaybe fromDynamic $ [toDyn "one", toDyn (\x -> x::Integer), toDyn 3, toDyn ()]

but it returned

["()"]

Answer 1

Your code does not do what you expect it to do. Long before the dynamic behaviour of Data.Dynamic kicks in, the Haskell type checker resolves the types. The type of the right part of the expression is

mapMaybe fromDynamic $ [toDyn "one", toDyn (\x -> x::Integer), toDyn 3, toDyn ()] :: Typeable b => [b]

and the type of the left part is

map show :: Show a => [a] -> [String]

so to combine these, the type variable b resp. a gets unified. If you were to compile this from a regular Haskell file, the compiler would give you a warning (The type variable `a' is ambigous). But in GHCi, the interpreter just defaults to ().

But this fixes the type of fromDynamic in the expression to Dynamic -> Maybe (), effectively selecting all elements of type ().

If you force the compiler to use a different type there, e.g. by specifying a type signature, you see that fromDynamic selects a different type:

Prelude Data.Dynamic Data.Maybe> map (show :: Integer -> String) . mapMaybe fromDynamic $ [toDyn "one", toDyn (\x -> x::Integer), toDyn 3, toDyn ()]
["3"]

Unfortunately, there is no way to achieve what you want: Select all elements whose type support a show instance, as that information is not available to fromDynamic.