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

[Daniel Frumin]

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

Exactly. 
x is of type Char and xs is of type [Char].
The list concatenation function (++) expects both of its arguments to be lists, so that's the reason you need to turn a Char x into a list containing only one value ([x]).



—
Sincerely yours,
Daniil Frumin



Angus Comber <anguscomber at gmail.com="mailto:anguscomber at gmail.com">> wrote:
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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