To check whether the given string is palindrome-method 2
by Ranjith | December 4th, 2008.Here is another program, advanced version of the previous program, to check if the entered string is a palindrome. Palindrome is a string segment, which reads same from both the directions. The same method is implemented here in the program also.
Logic : The given string is referred by two names, where one is traced by the SOL (Start Of Line) and the other is traced from EOL (End Of Line). In every iteration, the condition is checked, if both the characters are the same. If so, sets a flag, “pal” as true (i.e. 1), and hence prints out the result.
The problem can also solved in another method, where we check by reversing the string and then comparing.
#include<stdio.h>
#include<conio.h>
int stpal(char str[50]);
void main()
{
char str[50];
int pal;
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str);
pal = stpal(str);
if(pal)
printf(“\n\t THE ENTERED STRING IS A PALINDROME”);
else
printf(“\n\t THE ENTERED STRING IS NOT A PALINDROME”);
getch();
}
int stpal(char str[50])
{
int i = 0, len = 0, pal = 1;
while(str[len]!=’\0′)
len++;
len–;
for(i=0; i<len/2; i++)
{
if(str[i] == str[len-i])
pal = 1;
else
{
pal = 0;
break;
}
}
return pal;
}
Download exe and source code here.
October 10th, 2009 at 1:28 pm
It works fine but it is case sensitive. How to solve that?
January 17th, 2010 at 2:29 pm
THIS THING ALSO WORKS !! EASILY UNDERSTANDABLE !!!
#include
#include
int main()
{
int i,j,check=0;
char str[100];
printf(“ENTER A STRING:\n”);
gets(str);
for(j=0;str[j]!=”;j++);
j=j-1;
i=0;
while(i<=j)
{
if(str[i]==str[j])
{
i=i+1;
j=j-1;
check=1;
}
else
{
check=0;
break;
}
}
if(check==1)
printf("PALINDROME:");
else
printf("NOT A PALINDROME\n");
getch();
}
June 5th, 2010 at 11:47 am
i couldnt understand ur 9th number line// for(j=0;str[j]!=”;j++);
can u pls explain?
August 2nd, 2010 at 10:59 pm
i couldn’t understand the following part:
for(j=0;str[j]!=” “;j++)
j=j–;
i=0