Checking for palindrome string in c - c

Question

I am accepting a string as a command line argument. I want to check whether the inputted string is a palindrome or not and print the result. I have written the following code. But its displaying the result 'not palindrome' for all inputs.
#include<stdio.h>
#include<string.h>
int main(int argc, char argv[20]) {
int i;
int l = strlen(argv);
char str[20];
bzero(str, 20);
for(i=0; i<l; i++)
{
str[i] = argv[i+2];
}
int flag;
int len = strlen(str);
for(i=0; i< len/2; i++)
{
if(str[i] == str[len - (i+2)])
{
flag = 0;
}
else
{
flag = 1;
break;
}
}
if(flag == 0)
printf("Palindrome\n");
else
printf("Not a palindrome\n");
}

Answer 1

You could do it in a K&R-style by having two offset iterators in a for-loop:
#include <stdio.h>
#include <string.h>
#include <assert.h>
int main(int argc, char *argv[]) {
assert(argc != 1);
char *text = argv[1];
int len = strlen(text);
int is_palindrome = 1;
int i, j;
for(i = 0, j = len - 1; i < len / 2; i++, j--) {
if(text[i] != text[j]) {
is_palindrome = 0;
break;
}
}
printf("%s a palindrome.\n", is_palindrome ? "is" : "isn't");
return(0);
}
Changes from original:
Changed shift(len >> 1) to division(len / 2) as tenfour suggested.

Answer 2

Updated based on comments:
int is_palindrome(const char *s)
{
const char *t = s + strlen(s);
while (s<t && *s==*--t) s++;
return s>=t;
}
And since the OP wants a version that's not so heavy on pointers:
int is_palindrome(const char *s)
{
size_t i=0, j = strlen(s);
while (i<j && s[i]==s[--j]) i++;
return i>=j;
}
For reference, here's the original buggy version:
int is_palindrome(const char *s)
{
const char *t = s + strlen(s) - 1;
while (s<t && *s++==*t--);
return s>=t;
}

Answer 3

For one thing, your signature for main is off. It should be int main(int argc, char** argv) or int main(int argc, char * argv[]). You're treating a pointer to a string as if it were a string.
When you've changed that, the string you want should be in argv[1] (since argv[0] is some representation of the program name).

Answer 4

There's a good case for using pointers rather than indexes for this:
int is_palindrome(const char *s) {
const char *end = s + strlen(s);
while (end > s) {
--end;
if (*end != *s) return 0;
++s;
}
return 1;
}
If you like short, confusing code, you can re-write that:
int is_palindrome(const char *s) {
const char *end = s + strlen(s);
while (end > s) if (*(--end) != *(s++)) return 0;
return 1;
}
argv isn't a string, it's an array of strings, one for the program name and then one for each argument (usually space-separated in a command line). So to test if the first argument is a palindrome, you're interested in argv[1].
int main(int argc, char **argv) {
if (argc != 2) {
printf("usage: %s <string>\n", argv[0]); // or something
return 1;
}
if (is_palindrome(argv[1])) {
printf("Palindrome\n");
} else {
printf("Not a Palindrome\n");
}
}

Answer 5

The first loop doesn't make sense.
Copying the string to another doesn't make sense.
Just do it and adjust the index:
#include<stdio.h>
#include<string.h>
int main(int argc, char **argv) {
int i;
char * str = argv[1];
int flag;
int len = strlen(str);
for(i=0; i< (len+1)/2; i++)
{
printf("DEBUG: Comparing %c %c\n",str[i], str[len - (i+1)]);
if(str[i] == str[len - (i+1)])
{
flag = 0;
}
else
{
flag = 1;
break;
}
}
if(flag == 0)
printf("Palindrome\n");
else
printf("Not a palindrome\n");
}

Answer 6

No pointers (except the one use for making a copy of the original string).
#include <stdio.h>
#include <string.h>
int main( int argc, char *argv[] )
{
char *s2;
if ( argc != 2 )
return ( 1 ); // not properly invoked
if ( (s2 = strdup( argv[1] )) == NULL )
return ( 2 ); // failed (not likely)
printf( "\"%s\" %s a palindrome.\n", argv[1], strcmp( argv[1], strrev( s2 ) ) ? "is not" : "is" );
free( s2 );
return ( 0 );
}