Haskell: Pig Latin - haskell

Question

Question: The simple rule for translating into “Pig Latin” is to take a word that begins with a vowel and add "yay", while taking any word that be- gins with one or more consonants and transferring them to the back before appending "ay". For example, "able" becomes "ableyay" and "stripe" becomes "ipestray". Write a function that converts a string of letters into its Pig-Latin translation.
Implementation:
-- define function to detect vowel
isVowel :: Char -> Bool
isVowel c = elem c ['u','e','o','a','i']
-- define function Latin Pig
lp ::String -> String
lp str = if (isVowel (head str)) then do {str ++ "yay"}
else
do {
str ++ (head str)
tail str
lp str
}
Problem: So far I do not see any problem with my code (logic). Honestly this is my homework for an Intro to Haskell course. However the compiler gives me errors:
**Couldn't match expected type `t0 -> t1 -> t2 -> t3 -> [Char]'
with actual type `Char'
Expected type: [t0 -> t1 -> t2 -> t3 -> [Char]]
Actual type: String
In the first argument of `head', namely `str'
In the second argument of `(++)', namely
`(head str) tail str lp str'
Failed, modules loaded: none.**
What is wrong with my code?!

Answer 1

To begin, think about pattern matching.
Any not empty list can be define as x:xs, with,
x as head list
xs as tail list
then your code become,
-- define function Latin Pig
lp :: String -> String
lp [] = ""
lp (x:xs) = if (isVowel x) then str ++ "yay"
else ..... -- fill here
where str = (x:xs)
don't forget that the operator : is the constructor of list, for example,
'a':"bab" => "abab"
Keep in mind that a string is a list of char.
Furthermore, you can skip the where clause in the previous example, like this,
-- define function Latin Pig
lp :: String -> String
lp [] = ""
lp str#(x:xs) = if (isVowel x) then str ++ "yay"
else ..... -- fill here
Should be enough to help you.
Good luck

Answer 2

Here is another way of putting the pig latin rule:
if a word doesn't begin with a run of consonants, then the translation is the original word followed by "yay"
otherwise the translation is the rest of the word, followed by the initial run of consonants, followed by "ay"
This is easier to translate into Haskell.
(Actual Haskell code elided, as this is homework.)
You will find it helpful to combine your isVowel function with the break function from Data.List.

Answer 3

I'm not sure if it is part of your requirements to use recursion but here is my take on your task. You do not need to use the do monad to achieve what you want (unless that is an objective of the assignment?).
You might want to consider using pattern matching and guards instead of an if else block.
Also, like zurgl said, you can take advantage of matching the string like this: string#(x:xs) which will allow you to do work on the entire string while also using the head x and tail xs.
Note: All strings are lists.
Here is a quick example of what I suggested.
-- define function to detect vowel
isNotVowel :: Char -> Bool
isNotVowel c = notElem c ['u','e','o','a','i']
-- define function Latin Pig
lp :: String -> String
lp [] = []
lp p#(x:xs)
| not $ isNotVowel x = p++"yay"
| otherwise = let (constants, rest) = span isNotVowel p
in (rest++constants++"ay")
Have fun learning haskell!
Some nice resources for learning haskell:
Learn You A Haskell For Great Good
Real World Haskell

Answer 4

Kindly refer following code.
private static BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
System.out.print("Enter sentence: ");
String sentence = getString();
String latin = pigLatinForSentence(sentence);
System.out.println(latin);
}
private static String pigLatinForSentence(String s) {
String latin = "";
int i = 0;
while (i < s.length()) {
while (i < s.length() && !isLetter(s.charAt(i))) {
latin = latin + s.charAt(i);
i++;
}
if (i >= s.length())
break;
int begin = i;
while (i < s.length() && isLetter(s.charAt(i))) {
i++;
}
int end = i;
latin = latin + pigWord(s.substring(begin, end));
}
return latin;
}
private static boolean isLetter(char c) {
return ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'));
}
private static String pigWord(String word) {
int split = firstVowelInSentence(word);
return word.substring(split) + "-" + word.substring(0, split) + "ay";
}
private static int firstVowelInSentence(String word) {
word = word.toLowerCase();
for (int i = 0; i < word.length(); i++)
if (word.charAt(i) == 'a' || word.charAt(i) == 'e' || word.charAt(i) == 'i' || word.charAt(i) == 'o'
|| word.charAt(i) == 'u')
return i;
return 0;
}
private static String getString() throws IOException {
return buf.readLine();
}
Hope this solution help you learn more detail about pig latin.