C/C++ Code Snippet: Palindrome Function source code

/*   
      A palindrome is a word, phrase, number, or other sequence of symbols or elements, 
  whose meaning may be interpreted the same way in either forward or reverse direction.
  This program will prompt the user to enter a word or a sentence and then confirms if 
  the user's entry is an palindrome or not.   
*/
#include <iostream>
#include <cstdlib>
 
bool isPalindrome( char [] );
 
int main( int argc, char *argv[] )
{
    char userString[300];
    std::cout << \"Type a word or phrase to see if it's a Palindrome.\" << \"\n\n\";
    std::cin.getline(userString, 300);      
    std::cout << \"\n\";
    if ( isPalindrome( userString ) )
            std::cout << \"\\"\" << userString << \"\\" is a Palindrome.\";
    else
            std::cout << \"\\"\" << userString << \"\\" is not a Palindrome.\";   
    std::cout << \"\n\n\";
    system( \"PAUSE\" );        
    return 0;
}
/*
  The following function will return true if the array parameter is a palindrome.
   
  Step 1: First it counts how many characters are in the parameter array and then creates a new 
          array (array1) with that count size.
 
  Step 2: The contents of the parameter array are copied to array1 with no spaces or punctuations.
          - This is accomplished by checking the user's array for anything other than an 
            alphetbet character. When an non-alpha character is found, that array element 
            postion is recorded. It then copies the user's array contents from the starting 
            element position to the position right before the non-alpha character to the new
            array (array1). The element position right after the non-alpha character then 
            becomes the new starting position and the process iterates until array1 has only 
            one element position left in which a null terminator will be assigned.
 
  Step 3: array1 is then tested to see if it's a palindrome. Instead of copying array1 into an 
          another array in reverse order and then comparing the two, the test can be 
          accomplished by comparing array1 within itself. 
          - This is accomplished by comparing array1's first character with it's last character. 
            if both those charcters are the same, whether uppercase or lowercase, it then compares 
            array1's second character with it'a second-to-last character....and so forth until the 
            middle of array1 is reached. If at anytime, however, a comparison is not the same then 
            the 'isPalindrome' flag is set to false and the loop ends. The 'isPalindrome' flag will 
            remain true if and only if all comparisons are the same.
             
  Step 4: The function then returns the boolean value of isPalindrome.
*/
bool isPalindrome( char array[] )
{
    bool isPalindrome = true;
    int size = 0, index = 0, startingPos = 0, count1 = 0;   
    // Step 1
    while ( array[size] != '\0' ) 
    {                              
            size++;
    }   
    char array1[size + 1];
    // Step 2
    while ( index < size ) 
    {                     
            while ( (array[index] >= 'A' and array[index] <= 'Z') or
                    (array[index] >= 'a' and array[index] <= 'z')    )
                    index++;
            for ( int count = startingPos; count < index; count++ )
            {
                    array1[count1] = array[count];
                    count1++;
            }
            index++;
            startingPos = index;
    }
    array1[count1] = '\0';  
    //Step 3
    index = 0;
    while ( index <= (count1 - 1)/2 and isPalindrome )
    {
            if ( array1[index] != array1[count1 - index - 1]      and
                 array1[index] != array1[count1 - index - 1] - 32 and
                 array1[index] != array1[count1 - index - 1] + 32 and
                 array1[index] - 32 != array1[count1 - index - 1] and
                 array1[index] + 32 != array1[count1 - index - 1]     )
                    isPalindrome = false;
            index++;
    }
    // Step 4
    return isPalindrome;
}