使用指针检查字符串是否为回文

Question

What does the bold part stand for in this code? I've managed to fulfill most of the template (given by my teacher as assignment), can't find the reason behind that bold part.

I could use the strlen part before it appeared, I know one can still write the code without strlen at all.

#include <stdio.h>

int strlen(char *str)
{
    int c = 0;
    while (str[c] != '\0')
    {
        c++;
    }
    return c;
}

int isPalindrome(char *str)
{
    char *ptr1 = str;
    char *ptr2 = str + strlen(str) - 1;
    while (ptr2 > ptr1)
    {
        if (tolower(*ptr1) != tolower(*ptr2))
        {
            return (0);
        }
        ptr1++;

        ptr2--;
    }
    return (1);
}
/**
            This function will
                -return 1 if the given string is Palindrome,
                -return 0 if the given string is not Palindrome.
        */

**int n = strlen(str);
// Write your code here**
}

int main()
{
    char *str = "ABCCBA ABCCBA"; //your input
    if (isPalindrome(str))
    {
        printf("%s is a palindrome", str);
    }
    else
    {
        printf("%s is not a palindrome", str);
    }
}

Answer 1

The teacher hinted that using the length of the zero-terminated input string will be needed. They stored that useful value inside a variable.

If you are required to use the code provided by the teacher, then move all your code AFTER that line and within your code replace the strlen(str) with n .

int isPalindrome(char *str)
{
        /**
            This function will
                -return 1 if the given string is Palindrome,
                -return 0 if the given string is not Palindrome.
        */

    int n = strlen(str);
    // Write your code here

    char *ptr1 = str;
    char *ptr2 = str + n - 1;
    while (ptr2 > ptr1)
    {
        if (tolower(*ptr1) != tolower(*ptr2))
        {
            return (0);
        }
        ptr1++;

        ptr2--;
    }
    return (1);
}

Answer 2

I think that the function template starts from the declaration of the variable n that gets the size of the passed string.

int isPalindrome(char *str)
{
    /**
        This function will
            -return 1 if the given string is Palindrome,
            -return 0 if the given string is not Palindrome.
    */
    int n = strlen(str);
    // Write your code here*
}

But you ignored this declaration and used the expression strlen( str ) in this declaration

char *ptr2 = str + strlen(str) - 1;

instead of writing for example

int n = strlen(str);
//...
char *ptr2 = str + n - 1;

In any case the assignment looks not very well. For example this declaration

char *ptr2 = str + strlen(str) - 1;

can invoke undefined behavior when the passed string is empty.

The functions' parameters should be declared with the qualifier const because passed strings as arguments are not changed in the functions.

The argument of the standard function tolower must be converted to the type unsigned char .

The functions could look the following way

size_t strlen( const char *str )
{
    size_t n = 0;

    while ( str[n] ) ++n;

    return n;
}

int isPalindrome( const char *str )
{
    const char *ptr1 = str;
    const char *ptr2 = str + strlen( str );

    if ( ptr1 != ptr2 )
    {
        while ( ptr1 < --ptr2 && 
                tolower( ( unsigned char )*ptr1 ) == tolower( ( unsigned char )*ptr2 ) ) 
        {
            ++ptr1;
        }           
    }

    return !( ptr1 < ptr2 );
}

Also if the function strlen also must use pointers then it can look the following way

size_t strlen( const char *str )
{
    const char *p = str;

    while ( *p ) ++p;

    return p - str;
}

Here is a demonstration how your functions should look.

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

size_t strlen( const char *str )
{
    const char *p = str;

    while ( *p ) ++p;

    return p - str;
}

int isPalindrome( const char *str )
{
    const char *ptr1 = str;
    const char *ptr2 = str + strlen( str );

    if ( ptr1 != ptr2 )
    {
        while ( ptr1 < --ptr2 && 
                tolower( ( unsigned char )*ptr1 ) == tolower( ( unsigned char )*ptr2 ) ) 
        {
            ++ptr1;
        }           
    }

    return !( ptr1 < ptr2 );
}


int main(void) 
{
    char *str = "ABCCBA ABCCBA"; //your input

    if ( isPalindrome( str ) )
    {
        printf("\"%s\" is a palindrome\n", str);
    }
    else
    {
        printf("\"%s\" is not a palindrome\n", str);
    }   

    return 0;
}

The program output is

"ABCCBA ABCCBA" is a palindrome

Answer 3

You can implement isPalindrome without any knowledge about pointers. I prefer to use indices instead of pointers, look at the next example

#include <stdio.h>
#include <assert.h>

int strLen(const char *str)
{
    int n = 0;
    while(str[n]) { ++n; }
    return n;
}

int isPalindrome(const char *str) {
    int i = 0;
    int j = strLen(str) - 1;
    while (i < j) {
        if (str[i++] != str[j--]) { return 0; }
    }
    return 1;    
}

int main() {
    assert(isPalindrome(""));
    assert(isPalindrome("a"));
    assert(isPalindrome("aa"));
    assert(isPalindrome("aba"));
    assert(!isPalindrome("ab"));
    assert(!isPalindrome("aab"));
}