Quick contest (all languages)

[md]

Ok... so I gave the challenge on IRC, but I figure I'll open it to anyone on compsci

The Challenge: write a function to check to see if a string is a palendrome. It should only one parameter, the string, and return a boolean value of true if the string is a palendrome, or false if it is not.
[edit] You can optionally also pass a second parameter to specify the length of the string, just to reduce the overhead of recursion...

The Judges: I'll judge who wins by who can write the shortest code (instruction wise), who can write the fastest code (order n wise), and who's code is easiest to read. I might get a few other people (wtd, rizzix) to help judge too if I can.

The prize: the proze will be all my bits devided amongst the winners, or if there is a mod who will sponsor this however much they feel like giving.

[[Gandalf]]

code:

fcn checkPalindrome (word : string) : boolean     for i : 1 .. length (word) div 2         if word (i) ~= word (length (word) - i + 1) then             result false         end if     end for     result true end checkPalindrome

[md]

My C/C++ entry so people have an idea how one could do it

c++:

bool Palindrome(char *string, int length) {         return (length <= 0) || (string[0] == string[length-1]) && Palindrome(++string, length-2); }

[beard0]

Beat this :)

Turing:

fcn isPalindrome (str : string) : boolean     result length(str)<=1 or (str (1) = str (length (str)) and isPalindrome (str (2 .. length (str) - 1))) end isPalindrome

Hooray for Turing short circuit logic evaluation!

[beard0]

By the way Cornflake's "idea how one could do it" doesn't actually work, you need to expand on it.

[md]

Yes, yes... so my mind was a little slow after my exam... it's changed and it does now work, I've tested it.

[edit] arg... it's harder then it seems to write it on one line... see my sig
[edit 2] my god I'm dumb... but it now works... tested verilly...

[wtd]

O'Caml:

code:

let rec palindrome str =   let len = String.length str in     if len <= 1 then true     else if str.[0] = str.[len - 1] then       palindrome (String.sub str 1 (len - 2))     else false

[Martin]

For an added challenge:
Do it as a one liner recursive call to main in C, returning a number other than 0 if argv[0] is a palindrome, 0 if it isn't.

code:

int main (int argc; char**argv) {     return <yourcodehere>; }

[wtd]

Martin wrote:

For an added challenge:
Do it as a one liner recursive call to main in C, returning a number other than 0 if argv[0] is a palindrome, 0 if it isn't.

code:

int main (int argc; char**argv) {     return <yourcodehere>; }

Did you actually mean argv[0]? That's typically the executable name.

[Martin]

Sorry, yeah, I forgot about that. It's been a while since I've used C. argv[1];

[TokenHerbz]

To: Cornflake

you said you'd gimmy 5 bits on IRC for trying

When can i expect them?

[wtd]

Ruby:

code:

class String    def is_palindrome?       if length <= 1          true       elsif self[0] == self[-1]          self[1..-2].is_palindrome?       else          false       end    end end

[Tony]

Ruby:

class String   def is_palindrome?     self == self.reverse   end end

[md]

Methinks I'll leave this open until next weekend and then I'll judge winners... and martin; I've almost figured out how to beat your challenge

[goomba]

Python for the win!

code:

def isPalindrome(word):         if word[::-1] in word:                 return True         return False