Whether it is palindrome or not, recursively 2nd part

Well the idea came to me in the other post was more simple than I imagined, and now the recursive function itself looks much better, it's much more than a disguised cycle.

Extensively commented code, so here it is:

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# Include <string.h> # Include <iostream> using namespace std; int len, n = 0; chk4palindrosity string (string Thestra) { if (Thestra [0] == Thestra [Thestra. length () - 1]) / / compare first character with last { n + +; if (n == len / 2) / / if the number of times the comparison was true is ... return "If palindrome!" / / equal to half of the caracters is palindrome Thestra. erase (0, 1) / / delete first character Thestra. erase (thestr. length () - 1, 1) / / delete last chk4palindrosity return (Thestra) / / call the function with the string cut } else / / if one of the comparisons is true, is not a palindrome return "is not palindrome"; } int main () { inputf char [50] = {0}, input [50], * part; court <<"Enter a palindrome:"; cin. getline (input, '\ n'); part = strtok (input, "") / / strcat (inputf, part) / / while ((part = strtok (NULL, ""))! = NULL) / / strcat (inputf, part) / / remove spaces from array Thestra string (inputf) / / convert string array to facilitate operations len = Thestra. length () / / get the string LENGTH if (len == 1) court <<"If palindrome!" else court <<chk4palindrosity (Thestra) / / call function and print what return cin. get (); }

What makes the recursive function is to compare the first character to the last, if they are equal trim and knock: if not, it is palindrome. Any doubt already know that the comments are open.