将字符与 C 中的字符串进行比较

Question

I'm quite new to C and I'm wondering why in the code below, the char I'm comparing to each letter of the string word is showing that it's equal everytime. For example

If I've inputted the word

apple

and I'm looking for any repeating char in "apple" my function. I pass in to the function each char of apple such as a, p, p etc. It should return 1 when I pass in p since it's repeated, but instead, for every char of apple, my function says a == word[0], a == word[1] even though word[1] for "apple" is 'p'. I know char is ASCII, so each char has a number value, but I'm not sure why this is not working. Perhaps, I'm using the pointer *word in the functions arguments incorrectly?

My code is below for my function, rpt_letter:

int rpt_letter(char *word, char c)
{

    int i;
    int count = 0;
    i = 0;

    printf("This is the WORD %s\n", word);


    while(count < 2)
    {
        if(word[i] == c)
        {
            count++;
            printf("the count is %d\n the char is %c and the string is %c\n", count, c, word[i]);
        }
        i++;
    }

    if (count<2) 
    {
       // printf("letter %c was not found in the array. \n", c);
        return 0;
    } 
    else
    {
        //printf("letter %c was found at index %d in the array.\n", c, mid);
        repeats[rpt_counter] = c;
        rpt_counter++;
        return 1;
    }

    return 0;
}

I'll include the main method just in case -- but I believe the main method is working well

int main(void) 
{
    //! showArray(list, cursors=[ia, ib, mid])
    //int n = 51;
    char word[51]; 
    scanf("%s", word);

    //length of string
    for (n=0; word[n] != '\0'; n++); //calculate length of String
    printf("Length of the string: %i\n", n);

    int count = 0;

    //sort words

    int i;
    char swap = ' '; 

    for(int k = 0; k < n; k++)
    {
    for (i=0; i<n-1; i++)
    {

        //if prev char bigger then next char
        if (word[i] > word[i+1]) 
        {

            //make swap = prev char
            swap = word[i];

            //switch prev char  with next char
            word[i] = word[i+1];

            //make next letter char
            word[i+1] = swap;
        }
    }
    }
    printf("%s\n", word);

    for (i=0; i<n-1; i++) 
    {
     int rpt = rpt_letter(word, word[i]);  
     if(rpt == 1)
     {
         count++;
     }
    }

    printf("%d", count);

 return 0;   
}

I've tried a number of things such as using the operator !=, also <, > but it gives me the same result that each word[ia] == c.

Answer 1

You are getting this issue because in your code rpt_letter() the while loop has a terminating condition count >= 2 . Now consider input apple and character a . As a appears in apple only once, the count after traversing the whole word remains 1. But the loop doesn't terminate. So, the index i becomes greater than the length of string and tries to check the character appearing after that.

The loop terminates eventually when it gets another a this way. You need to add a check for the terminating null character in your loop so that it doesn't cross the length of the string .

Change the while loop condition to something like - while((count < 2) && (word[i] != '\\0'))