Exercise on Strings

Hello everyone,
i have to program an C++ program that will determine whether a given string is a palindrome or not.(A palindrome is a word that read same forward and backward...like "MADAM")

Here is my program, but i got some error and i don't know why?
Please help.

You have a lot of errors in that program syntactic and logical. Here's the fixed program :-)

Code:

#include <iostream.h>
#include <string.h>

int main()
{
	int length;
	int arr1=0;
	typedef char string[30];
	typedef char string2[30];
	string word[10];
	string2 invWord[10];
	cout<<"Enter word"<<endl;
	cin>>word[arr1];
	
	
	while (stricmp(word[arr1], "Quit")!= 0)
	{
		length=strlen(word[arr1]);
		
		for (int j=0, i=length-1; j<length; j++, i--)
		{
   			invWord[arr1][j]=word[arr1][i];
   		}

		invWord[arr1][length] = '\0';

		if (strcmp(word[arr1], invWord[arr1]) == 0)
   			cout<<"Palindrome"<<endl;

   		cout<<"Enter word"<<endl;
   		cin>>word[++arr1];
	}

   	return 0;
}

But you can do a lot better, here's my stab at it :-)

Code:

#include <functional>
#include <algorithm>
#include <iostream>
#include <string>

using namespace std;

class good_char: public unary_function<char, bool> {
  const string good;
public:
  good_char()
    : good("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
  {}

  bool operator()(const char c) const
  {
    if (good.find(c) != string::npos)
    {
      return true;
    }

    return false;
  }
};

bool is_palindrome(string& s)
{
  string::iterator new_end = remove_if(s.begin(), s.end(), not1(good_char()));
  s.erase(new_end, s.end());

  string r = s;

  reverse(r.begin(), r.end());

  if (s == r)
  {
    return true;
  }

  return false;
}

int main()
{
  string s;

  cout<<"Enter a word or phrase: ";

  getline(cin, s);

  if (is_palindrome(s))
  {
    cout<<"You entered a palindrome"<<endl;
  }
}

hey cela,
just wanted to make things clear in my head . What does these do?

Code:

class good_char: public unary_function<char, bool> {
  const string good;
public:
  good_char()
    : good(" abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWX
YZ")
  {}

  bool operator()(const char c) const
  {
    if (good.find(c) != string::npos)
    {
      return true;
    }

    return false;
  }
};

>>What does these do?
It's a function object that checks a character given to it with a list of valid characters. If it doesn't match one of those characters then it returns false and remove_if() deletes the character, just what we want. :-)

and hey cela ,
sorry again me here. what does this do??

Code:

string::npos
and lastly why is this empty function?
good_char()
    : good("  abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWX

YZ")
  {//why is it empty here?}

hi cela,

unfortunately, i didn't start the true 2 dimensional array yet("array [] []").

So, can you change the program without using the "real" 2 dimensional array?

Code:

#include <iostream.h>
#include <string.h>

int main()
{
  int length;
  typedef char string[30];
  typedef char string2[30];
  string word;
  string2 invWord;
  cout<<"Enter word"<<endl;
  cin>>word;

  while (stricmp(word, "Quit")!= 0)
  {
    length=strlen(word);

    for (int j=0, i=length-1; j<length; j++, i--)
    {
      invWord[j]=word[i];
    }

    invWord[length] = '\0';

    if (strcmp(word, invWord) == 0)
       cout<<"Palindrome"<<endl;

    cout<<"Enter word"<<endl;
    cin>>word;
  }

  return 0;
}

Thanks Cela,

but i don't understand this line:

invWord[length] = '\0'

Without this line, the program doesn't work, why?
if "invWord" need this line, how come "word" don't need this line?

Thanks

> invWord[length] = '\0'

This puts a null terminator at the end of the string - it tells the computer the string's done. word doesn't need one because it has one appended to it when it's read in in the cin statement.

Just wanted to toss in my 2 cents. Obviously it's going to have a lot less overhead as I've excluded any use of the stl, and it's only 4 lines long

Code:

bool isPalindrome(const char *str) {
  for (int l = 0, r = strlen(str) - 1; l < r; ++l, --r)
    if (str[l] != str[r])
      return false;
  return true;
}

>>Obviously it's going to have a lot less overhead as I've excluded any use of the stl, and it's only 4 lines long
It also doesn't have much flexibility. Since you're making an exact comparison the following palindrome returns false

Code:

a man, a plan, a canal, panama

Oh, yeah. You're right. I didn't take that into consideration. How about this then:

Code:

bool isPalindrome(const char *str) {
  for (int l = 0, r = strlen(str) - 1; l < r; ++l, --r) {
    while (!isalpha(str[l]))
      ++l;
    while (!isalpha(str[r]))
      --r;
    if (l < r && toupper(str[l]) != toupper(str[r]))
      return false;
  }
  return true;
}