[Haskell] Haskell [x] and x notation - as-pattern example

[Chernin, Nadav]

Hi,
Write firstLetter all@(x:xs) = "The first letter of " ++ all ++ " is " ++ [x] ++ " otherbit " ++ xs
Nadav


From: haskell-bounces at haskell.org [mailto:haskell-bounces at haskell.org] On Behalf Of Angus Comber
Sent: Wednesday, April 03, 2013 12:01 PM
To: Haskell Mailing List
Subject: [Haskell] Haskell [x] and x notation - as-pattern example

I am reading Learn you a Haskell for great good and on page 40 - as-patterns.

I have changed the example slightly to be:

firstLetter :: String -> String
firstLetter "" = "Empty string, oops"
firstLetter all@(x:xs) = "The first letter of " ++ all ++ " is " ++ [x] ++ " otherbit " ++ xs
Then can use like this:

*Main> firstLetter "Qwerty"
"The first letter of Qwerty is Q otherbit werty"
But I was confused about the difference between [x] and x and why I have to use [x] in the above example.

For example if I change to

firstLetter :: String -> String
firstLetter "" = "Empty string, oops"
firstLetter all@(x:xs) = "The first letter of " ++ all ++ " is " ++ x ++ " otherbit " ++ xs
I get error:

Couldn't match expected type `[Char]' with actual type `Char'
In the first argument of `(++)', namely `x'
In the second argument of `(++)', namely `x ++ " otherbit " ++ xs'
In the second argument of `(++)', namely
  `" is " ++ x ++ " otherbit " ++ xs'
I can use xs to print "werty" but have to use [x] to print "Q". Why is that?

What does [x] mean?

In the (x:xs) : just delimits each element. so x is the first element. Why can I not print by using x?

Also xs is of what type? list of values? So does this mean x is an element and xs must be of type list? Confused...

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