Read instances tend to be implemented such that they parse a string that looks like the source code for that type. This is usually the inverse of Show.

λ> show "hello"

"\"hello\""

λ> read "\"hello\"" :: String

"hello"

λ> read (show "hello") :: String

"hello"

On Fri, Dec 12, 2014 at 11:44 PM, Venu Chakravorty <venuchv@gmail.com> wrote:

Hello everyone,

I am new to Haskell and this might seem very naive, please bear with me.

=======================================
Prelude> read "4" + 4
8
Prelude> (read "4" :: Int) + 4
8
Prelude> read "hello " ++ "world"
"*** Exception: Prelude.read: no parse
Prelude> (read "hello" :: String) ++ " world"
"*** Exception: Prelude.read: no parse
=======================================

Could someone please explain why the last two statements don't work?

My understanding is that "read" has a type of "read :: (Read a) => String -> a".
So, "read "hello" " should give me an instance of type "Read" to which I am
appending the string "world" (just like the first 2 cases where I get
an instance of
"Read" ("Int" in this case) to which I am adding another "Int" and I
get a "Num" which
is then displayed). I expected to see "hello world" as the output.

Is it that the type "String" is not an instance of type class "Read"?
Please tell me what
I am missing here.

Regards,
Venu Chakravorty.
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