Q:

"read"-ing string data into haskell "data" types

Thanks to this excellent tutorial, I know how to read a string (in this case read from a file at people.txt directly into a type synonym:
type Person = [Int]

like this:
people_text <- readFile "people.txt"
let people :: [Person]
    people = read people_text

What I want to do is use a datatype (instead of a type synonym). 
Any pointers on what I am missing here?  I thought I would be able to read string-data directly into a Person -  defined like this (credit to learnyouahaskell.com)
data Person = Person String String Int Float String String deriving (Show)

When I try the obvious
 txt <- readFile "t.txt" (this works OK)

with t.txt containing
"Buddy" "Finklestein" 43 184.2 "526-2928" "Chocolate"

I get this error:

No instance for (Read Person)

A:

First of all, you need to derive Read for your type. 
You can think of read and show as being opposites, and a sort of poor man's serialization.  show lets you convert to String, read converts from String, and in most cases the String produced should also be valid Haskell code that, when compiled, produces the same value that read gives you. 
On that note, the contents of your file aren't going to work, because that's not the format used by the default implementations of read and show, i.e. the implementations you get by putting Read and Show in the deriving clause. 
For example, given this:
data Person = Person String String Int Float String String deriving (Read, Show)

buddy = Person "Buddy" "Finklestein" 43 184.2 "526-2928" "Chocolate"

Then in GHCi, we get:
> show buddy
"Person \"Buddy\" \"Finklestein\" 43 184.2 \"526-2928\" \"Chocolate\""

The quotes are escaped because that's a String value.  In the file, it would look like this:
Person "Buddy" "Finklestein" 43 184.2 "526-2928" "Chocolate"

Which you'll note is the same as the original definition in the source file. 

A:

Just add Read to the deriving
 data Person = Person String String Int Float String String deriving (Show, Read)

