C。 使用 3 个函数确定字符串是否为回文

Question

I'm supposed to make a code that determines if entered string is a palindrome, the code is supposed to have 3 functions. the first is to determine if sting is palindrome, the second is to use isalpha() to remove all non letters and the third is to use tolower to remove the difference between upper and lower case letters. I made all 3 separate functions but now that I have to combine them the code just gives exception thrown.

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

int isPalindrome(char inputString[])
{
    int middle = strlen(inputString) / 2;
    int len = strlen(inputString);
    for (int i = 0; i < middle; i++)
        if (inputString[i] != inputString[len - i - 1])
        {
            return 0;
        }
        else
        {
            return 1;
        }
}

void checker(char stringWithoutSpace[], char bigArray[]) 
{
    int j = 0;
    for (int i = 0; i < strlen(stringWithoutSpace); i++)
    {
        if (isalpha(stringWithoutSpace[i]) != 0)
        {
            bigArray[j++] = stringWithoutSpace[i];
        }
    }
} 

void lowerCase(char* string)
{
    int i;
    for (i = 0; string[i]; i++) 
    {
        string[i] = tolower(string[i]);
    }
}

#define SIZE 1000

int main(void) {
    int repeat = 1;
    char arrayPalindrome[SIZE];
    char bigArray[SIZE];
  
    while (repeat == 1)
    {   
        printf("Enter a sentence: ");
        scanf_s("%s", &arrayPalindrome);

        checker(arrayPalindrome, bigArray);
        lowerCase(arrayPalindrome, bigArray);

        if (isPalindrome(arrayPalindrome) == 1) 
        {
            printf("This is a palindrome.");
        }
        else
        {
            printf("this in not a palindrome: ");
        }
        printf("Do you want to enter another sentence (0 for no, 1 for yes)?");
        scanf_s("%d", &repeat);
    }

    return 0;
}

Answer 1

For starters this call of scanf_s is incorrect.

scanf_s("%s", &arrayPalindrome);

You have to write

scanf_s("%s", arrayPalindrome, rsize_t( sizeof( arrayPalindrome ) ) );

The function checker does not build a string in the array bigArray because you forgot to append the stored sequence of characters in the array with the terminating zero character '\0' .

Moreover the function does not make sense because the array bigArray is not used anymore in the program.

The function lowerCase is declared with one parameter

void lowerCase(char* string);

but is called with two arguments.

lowerCase(arrayPalindrome, bigArray);

Pay attention to that there is no need to declare the auxiliary array bigArry .

And the program should not change the original string.

Answer 2

Others, in comments and answers, have addressed the shortcomings in the code you presented. This answer addresses the heuristic approach to the problem.

As you are experiencing, every line of code is an opportunity for a bug to occur.

Acknowledging that the problem statement may have specified "write three functions", there is no need for so many. A single function does all that is necessary:

#include <stdio.h> // all that is needed

bool isPalindrome( const char s[] ) {
    // a "custom" 7bit ASCII "look up" table with exclusion ('.') and 'tolower()' equivalence
    char *tbl =
        "................................................0123456789......"
        ".abcdefghijklmnopqrstuvwxyz......abcdefghijklmnopqrstuvwxyz.....";

    // 'L'eft and 'R'ight indexing into the passed string
    size_t L = 0, R = 0;

    // find the end of the string (its 'r'ight index)
    while( s[R] ) R++; // poor boy's 'strlen(s)'. '\0' does not matter.

    do {
        // adjust indexes L & R rejecting characters
        // seek to "meet in the middle" of a palindromic string
        while( L <= R && tbl[ s[L] ] == '.' ) L++;
        while( L <= R && tbl[ s[R] ] == '.' ) R--;
    } while( L <= R && tbl[ s[L] ] == tbl[ s[R] ] && (L+=1) > 0 && (R-=1) > 0 );
    // string is palindromic if it has passed the tests above
    return L >= R;
}

int main() {
    // without the drudgery of re-entering test strings,
    // this "test harness" examines 3 different strings
    // giving the evaluation of each. Easy to add more test cases.
    char *strs[] = {
        "LeVrEl",
        "level",
        "123Madam I'm    Adam321",
    };

    for( int i = 0; i < sizeof strs/sizeof strs[0]; i++ )
        printf( "%s - '%s'\n", isPalindrome( strs[i] ) ? "Yes" : "No ", strs[i] );

    return 0;
}

No  - 'LeVrEl'
Yes - 'level'
Yes - '123Madam I'm    Adam321'

Far fewer lines of code, and should be "easy to follow" when one envisions a string as simply a contiguous array of characters.

If the assignment is to write 3 functions, one could add two more "no-op" functions to fulfill that criteria:

int thing1(void) { return 1; }
int thing2(void) { return 2; }