检查回文字符串的 C 程序

Question

I wrote two sample programs to check for a palindrome string. But in both I am getting output like, its not a palindrome number. What I am missing?

I strictly assume somehow code is executing my if statement and put flag in to 1. May be because of that length calculation. Anyone has a better idea?

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <conio.h>
    
    int main(void) {
        setbuf(stdout,NULL);
        char name[100];
        int i,length,flag=0,k;
        printf("Enter your name");
        /*scanf("%s",name);*/
        gets(name);
        length=strlen(name);
        for(i=0;i<=length-1;i++)
        {
            for(k=length-1;k>=0;k--)
            {
                if(name[i]!=name[k])
                {
                    flag=1;
                    break;
    
                }
    
                }
            }
    
        if(flag==0)
        {
            printf("Give word is a palindrome");
        }
        if(flag==1)
        {
            printf("This is NOT a palindrome word");
        }
        return 0;
        }

and

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <conio.h>

int main(void) {
    setbuf(stdout,NULL);
    char name[100];
    int i,length,flag=0;
    printf("Enter your name");
    /*scanf("%s",name);*/
    gets(name);
    length=strlen(name);
    for(i=0;i<=length/2;i++)
    {
        if(name[i]!=name[length-1])
        {
            flag=1;
        }
    }


    if(flag==0)
    {
        printf("Give word is a palindrome");
    }
    if(flag==1)
    {
        printf("This is NOT a palindrome word");
    }
    return 0;
    }

Answer 1

First Algorithm

The algorithm you are using in the first program involves comparing each letter to every other letter which does not help in determining if the number is a palindrome and it does not seem fixable.

Second Algorithm

The problem with the second approach, however, is you are always comparing name[i] to name[length] . Instead change it to length-i-1 . This will start comparing from length-1 and decrement the length of the character by 1 for every next iteration:

for(i = 0;i <= length / 2;i++)
{
    if(name[i] != name[length-i-1])
    {
        flag=1;
        break;
    }
}

gets() and buffer overflow

Do not use gets . This method is susceptible to a buffer overflow. If you enter a string longer than 100 characters, it will result in undefined behavior. Use fgets instead for deterministic behavior:

fgets(name, sizeof(name), stdin);

This takes in the size of the buffer and only reads up to sizeof(name) characters.

Full code

Ideally, you should consider wrapping the logic to check if the string is a palindrome in a function:

int is_palindrome(char*);

int main(void) 
{
    char name[100];
    setbuf(stdout,NULL);
    printf("Enter your name");
    fgets(name, sizeof(name), stdin);
    
    if(is_palindrome(name))
    {
        printf("The given word is a palindrome");
    }
    else
    {
        printf("This is NOT a palindrome word");
    }
    return 0;
    
}

int is_palindrome(char* name)
{
    int length = strlen(name);
    int flag = 0, i;
    for(i = 0;i <= length / 2; i++)
    {
        if(name[i]!=name[length-i-1])
        {
            return 0;
        }
    }
    return 1;
}

Answer 2

Try this

    #include <stdio.h>
    #include <stdlib.h>

    #include <string.h>
    int main() {
       char text[100];
       int begin, middle, end, length = 0;
       printf("enter the name: ");
       scanf("%s",text);
       while ( text[length] != '\0' ){
            length++;}

       end = length - 1;
       middle = length/2;
       for ( begin = 0 ; begin < middle ; begin++ ) {
           if ( text[begin] != text[end] ) {
              printf("Not a palindrome.\n");
              break;
           }
           end--;
         }
         if( begin == middle )
              printf("Palindrome.\n");
              return 0;
     }

The problem with the first piece of code is you are comparing it more than required, compare it with length-i-1 .

The main problem with the second code is you are comparing it with only the last letter of a word.

Hope you understood your mistake

Answer 3

There is plenty wrong with both your attempts. I strongly suggest using a debugger to investigate how your code works (or doesn't).

Your first attempt performs length ² (incorrect) comparisons, when clearly only length / 2 comparisons are required. The second performs length / 2 comparisons but the comparison is incorrect:

name[i] != name[length-1] ;

should be:

name[i] != name[length - i - 1] ;

Finally you iterate exhaustively when you could terminate the comparison as soon as you know they are not palindromic (on first mismatch).

There may be other errors - to be honest I did not look further than the obvious, because there is a better solution.

Suggest:

#include <stdbool.h>
#include <string.h>

bool isPalindrome( const char* str )
{
    bool is_palindrome = true ;
    size_t rev = strlen( str ) - 1 ;
    size_t fwd = 0 ;
    
    while( is_palindrome && fwd < rev )
    {
        is_palindrome = (str[fwd] == str[rev]) ;
        fwd++ ;
        rev-- ;
    }

    return is_palindrome ;
}

In use:

int main()
{
    const char* test[] = { "xyyx", "xyayx", "xyxy", "xyaxy" } ;
    for( size_t t = 0; t < sizeof(test)/sizeof(*test); t++ )
    {
        printf("%s : %s palindrome\n", test[t], 
                                       isPalindrome( test[t] ) ? "Is" : "Is not" ) ;
    }
    
    return 0;
}

Output:

xyyx : Is palindrome
xyayx : Is palindrome
xyxy : Is not palindrome
xyaxy : Is not palindrome