I am a beginner and I was trying to write a function that check if a string can be interpreted to number or not. Here is my code:
string' xs = if (all isDigit xs == False)
then "can not be interpreted"
else read xs::Int
But it keeps reporting the error " Couldn't match expected type ‘[Char]’ with actual type ‘Int’ " I don't know why, has someone ever met this problem?
Both branches of your if-then-else need to have the same type. Your "then" branch has type [Char] and your "else" branch has type Int. It seems like your "then" branch should result in some sort of error. In this case you can use error, which has a polymorphic type and can be used instead.
A better solution (suggested in the comment section) would be to use the Either type, which can return one of two options (Left or Right).
string' xs = if (all isDigit xs == False)
then Left "can not be interpreted"
else Right (read xs::Int)
Another common thing to do would be to use the Maybe type
string' xs = if (all isDigit xs == False)
then Nothing
else Just (read xs::Int)
string'- is itstring' :: String -> Stringorstring' :: String -> Int? In haskell, you can't have a function that returns one of two different types based on runtime properties of the input. – Daniel Martinreads :: String -> [(Int, String)], which will check that the input is well-formed for you, returning[]if not (and a list with the successful parse and whatever was left unparsed if so). – Daniel Wagnernot (...)instead of(...) == False. This is a matter of taste, but the first looks more "natural". Alternatively, remove that completely and swap the then/else branches. (Daniel's suggestion above aboutreadswould be even better.) – chi