Problem with the palindrome

hi
can anyone help me out of this program


/*program to determine a palindrome*/
#include<stdio.h>
#include<conio.h>
#include<string.h>

int main()
{
int check(char *start);
char word[40];
int result;

printf("\nEnter the string\t");
scanf("%s",word);

result=check(word);

if(result==1)
printf("\n%s is a palindrome\n",word);
else
printf("\n%s is not a palindrome\n",word);


getche();
return 0;
}

int check(char *start)
{
char *end;
int i,len=0;

while(*start!='\0')
{
len++;
*start++;
}
start=start-len;
end=start+len;

for(i=0;i<len/2;i++)
{
if(*start==*end)
{
*start++;
*end--;
return 1;
}
else
return 0;
}
}

You complicated this program too much.

Use this...

int Palindrome(char String[100])
{
int counter1=0, counter2=0, counter3=0;

counter2 = strlen(String);
for (counter1=0;counter1<=strlen(String); counter1--)
{
counter2--;
if (String[counter1] == String[counter2])
{
counter3 = 1;
}
}
return counter3;
}

It's easy. The string is read from the beginning to the end and reverse.

Use this as a template for your program. Since this is only one solution, see if you can come up with another.

Code:

#include <stdio.h>
#include <ctype.h>

int main(void){
	char string[1024];
	int tag, count, back, flag, loop = 1;

	while(loop == 1){
		flag = 1;

		puts("Enter a word, phrase, or sentence. Or type 'END' to exit:");
		for(count = 0; (string[count = getchar()) != '\n'; ++count)
			;
		if((toupper(string[0]) == 'E') && (toupper(string[1]) == 'N') &&
			(toupper(string[2]) == 'D'))
			break;
		tag = count - 1;

		for((count = 0, back = tag); count <= tag/2; (++count, --back)){
			if(string[count] != string[back]){
				flag = 0;
				break;
			}
		}

		for(count = 0; count <= tag; ++count)
			putchar(string[count]);
		if(flag == 1)
			puts(" is a palindrome\n");
		else
			puts(" is not a palindrome\n");
	}
}

-Prelude

>> if((toupper(string[0]) == 'E') && (toupper(string[1]) == 'N') &&
(toupper(string[2]) == 'D'))
<<

Optimized:

Code:

strupr(string);

if (!strcmp("END", string))
/*** ETC ***/

strupr(string);

if (!strcmp("END", string))
/*** ETC ***/



Does not seem to work...FYI

Mattz

Let me see all of your code. Don't forget to use the code tags to make it readable.

Here is my first attempt. Modified it a bit and it seems to work..but noticed its probably not in the right place as it should print out the contents of buffer when I type QUIT instead of just abruptly ending.
<code>
#include <stdio.h>
#include <string.h>

int main()
{

/* A struct is used to define the character
array */

struct str_array{
char buffer[50];
};



int i; /* to be used as an index */

char in_val[50]; /* used within gets() */

struct str_array buf[10]; /* an array of
struct str_array

buf[0] .. buf[9] */

/* as a demo, prompts for seven strings are
provided */



for (i=0; i<=6; i++)
{

printf("Enter a string -> ");
fflush(stdin); /* definitely, for gets() */
gets(in_val); /* input string */

/* syntax of strcpy(a,b) is such that a
is the "destination" and b is the
"source" */

strcpy(buf[i].buffer, in_val); /* note
buf[i].buffer */

/* some light testing is performed */

printf("Just after gets now!\n");
printf("String provided as input was: ");
puts(in_val);

if (strcmp("QUIT", in_val)==0)

return 0;


}

/* contents of the array of structs
provided as output */

for (i=0; i<=6; i++)
{

puts(buf[i].buffer);

}

return 0;

}
</code>

This should help:

#include<stdio.h>
#include<stdlib.h>
int main()
{
char string1[20];
puts("Enter string one.");
gets(string1);
char string2[20];
string2=strrev(string1);
if (strcmp(string1,string2)==0)
{
puts("Palindrome.");
}
else
{
puts("Not a palindrome.");
}
return 0;
}
Just an idea for a syntax. Cya.

char string2[20];
string2=strrev(string1);

No. This is incorrect. You cannot use an array to handle a char* return value like this. 'string2' must be a 'char*'.

Quzah.

QUZAH,
Any ideas on how to fix this program I posted earlier? Want to have the user type "QUIT" to exit and then print out the string arrays.

Mattz

or if they actually fill the 7 arrays that are counted it prints.