Thread: palindrom using recursion?

Code:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int is_palindrom(char niz[]);

int main(int argc, char *argv[])
{
  if(argc<2)
    { 
      printf("Need an argument-a word!\n");
      exit(1);
    }

  if(strlen(argv[1])==0 || strlen(argv[1])==1)
    { printf("You have to input a word ,not a character!\n");
    exit(1);
    }

  if(is_palindrom(argv[1])==0)
    {
      printf("A word %s is palindrom!\n",argv[1]);
    }

  else printf("A word %s is not a palindrom!\n",argv[1]);
 
}

int is_palindrom(char niz[])
{
  char reverse_niz[strlen(niz)];
  int i;
  int x=0;
  for(i=strlen(niz)-1;i>=0;i--)
    {
      reverse_niz[x]=niz[i];
      x++;
    }
  reverse_niz[x]='\0';
     
  if(strcmp(reverse_niz,niz)==0)
    return 0;

  else 

    return 1;

}

The above program works. A palindrom is a a word which can be read backwards the same as forward(ex. cepec). In the above program i first reversed the inputed word and stored it into reverse_niz and then compare that string to the original. And if the words are the same the thing is palindrom.

BUT I have to implement that function is_palindrom(char niz) using recursion. Somehow I have to make for the method to be callig itself in the body of a method.

...what a dounting task

so if n is 3 the output would be 6?

no if n is 3 it would print
1 2 3

Code:

 int count (int n) {
    if(n>=1)
    return n - count(n-1);
    }

Maybe like this?
Any good tutorials about recursion,cause this is a bit new to me.

thanks micko!

I am on it. Ill be back!

btw: Thantos, I meant the factorial function. The result is 6 if n =3.

I get a segmentation fault error on this one. Im trying to do that count thing from 1 to n. It looks ok to me . What did I do wrong?

Code:

#include<stdio.h>
int count(int n);
int main()


 {
   int z=3;
   printf("%d\n", count(z));

   return 0;
 }

int count(int n) {
  int i=1;
  while(i<=n) {
    return  count(n)-count(n-i);
    i++;
  }
  return 1;
 

}