CODEWITHSUNDEEP
haskell
Herokiller
Herokiller

Reputation: 2991

Haskell Prelude.read: no parse String

from haskell examples http://learnyouahaskell.com/types-and-typeclasses

ghci> read "5" :: Int  
5  
ghci> read "5" :: Float  
5.0  
ghci> (read "5" :: Float) * 4  
20.0  
ghci> read "[1,2,3,4]" :: [Int]  
[1,2,3,4]  
ghci> read "(3, 'a')" :: (Int, Char)  
(3, 'a')

but when I try

read "asdf" :: String

or

read "asdf" :: [Char]

I get exception

Prelude.read No Parse

What am I doing wrong here?

Upvotes: 39

Views: 40808

Answers (1)

bheklilr
bheklilr

Reputation: 54078

This is because the string representation you have is not the string representation of a String, it needs quotes embedded in the string itself:

> read "\"asdf\"" :: String
"asdf"

This is so that read . show === id for String:

> show "asdf"
"\"asdf\""
> read $ show "asdf" :: String
"asdf"

As a side note, it's always a good idea to instead use the readMaybe function from Text.Read:

> :t readMaybe
readMaybe :: Read a => String -> Maybe a
> readMaybe "asdf" :: Maybe String
Nothing
> readMaybe "\"asdf\"" :: Maybe String
Just "asdf"

This avoids the (in my opinion) broken read function which raises an exception on parse failure.

Upvotes: 52

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