定义返回类型IO [a] Haskell

Question

I want to return the expected type IO [a]

I tried: Type that I want

fc :: [a] -> IO [a]
fc a = sequence $ map show list

> fc [1,2,3]
> [1,2,3]

What can I do?

Answer 1

If you have a pure value, the way to convert to into a monadic value is with the return (aka pure ) function:

fc :: [a] -> IO [a]
fc list = return list

But if you want to map the show function over your list, the result will be [String] , because the function show :: a -> String turns any value into a String , so:

fc :: Show a => [a] -> IO [String]
fc list = return $ map show list

Notice the Show a constraint. It tells the compiler that the type a , whatever it is, must support function show . Without it, such function won't type check.

The sequence function is beside the point completely, since it turns a list of monadic values into one monadic value that is a list. If you wanted to use that function, then the list you pass to it must be a list of monadic values, for example:

fc :: Show a => [a] -> IO [()]
fc list = sequence $ map print list

Here, print is an IO () action that prints out a value.

If you want a more precise answer, you'd need to specify more clearly what you're trying to do.