Hi Russ, You probably have the monomorphism restriction turned of, so you don't get an error message. If you add a type, it will work fine. Like this: Prelude> read "123" *** Exception: Prelude.read: no parse Prelude> read "123" :: Int 123 Prelude> Greets, Edgar On Wed, Oct 27, 2010 at 8:36 AM, Russ Abbott wrote:

Hi,

I'm having a problem with the read function. I thought it took a string and converted it into a value (if there is a conversion). But it doesn't seem to work for me on GHCi

Prelude> read "123" *** Exception: Prelude.read: no parse Prelude> read (show 123) *** Exception: Prelude.read: no parse

There must be something trivial that I'm missing.

Thanks.

*-- Russ Abbott*

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

On Tue, Oct 26, 2010 at 11:36:35PM -0700, Russ Abbott wrote:

Hi,

I'm having a problem with the read function. I thought it took a string and converted it into a value (if there is a conversion). But it doesn't seem to work for me on GHCi

Prelude> read "123" *** Exception: Prelude.read: no parse Prelude> read (show 123) *** Exception: Prelude.read: no parse

Since read has type read :: Read a => String -> a read "123" has type read "123" :: Read a => a that is, read "123" can have *any* type which is an instance of Read. GHC cannot figure out which type you want just by looking at the string. Now, normally in this situation you would get an "ambiguous type" error. However, ghci also has sometype defaulting rules: normally they make things easier/more intuitive, but in this particular case they make things more confusing! What is happening is that since ghci doesn't know what type your expression should have, it defaults it to type (). Now when it tries to read "123" as a value of type () it fails. As someone else already pointed out, the solution is to give a type annotation to let ghci know what type you are expecting. Prelude> read "123" :: Int 123 Prelude> read "123" :: Double 123.0 Note, however, that when using read in a the larger context of a program, it is usually unnecessary to provide a type annotation, since the type can be inferred from the context. For example, addStrings :: String -> String -> Int addStrings x y = read x + read y -Brent

There must be something trivial that I'm missing.

Thanks.

*-- Russ Abbott*

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners