How does character array behave when initialized via strcpy?

Question

I am writing a simple code to find a string s1 from a string s2 and then saving it in string s3. The function returns 1 if the string s1 is present in s2 else 0. I am using pointers as parameters. The problem is that even when I am assigning s1 to n, line 30 of code isn't providing me correct output. Also, the code is behaving abnormally. Can you please point out the error?

#include<string.h>
#include<stdio.h>
int code(char * s2, char * s1, char * s3)
{
    char h[] = "";
    strcpy(h, s2);
    printf("%s\n", h);

    char n[] = "";
    strcpy(n, s1);
    printf("%s\n", n);

    char b[] = "";

    int i = 0, j = 0, flag = 0, flag2 = 0;

    for (i = 0; n[i] != '\0'; i++)
    {
        printf("%d\n", i);
        printf("%c\n", n[i]);
        for (j = 0; h[j] != '\0'; j++)
        {
            printf("%c\n", h[j]);
            if (n[i] == h[j])
            {
                flag = 1;
                break;
            }

        }
        if (flag == 1)
        {
            break;
        }
    }

    printf("%d\n", i);
    printf("%d\n", j);

    printf("%c\n", n[i]);
    printf("%c\n", h[j]);

    printf("\n");

    if (flag == 1)
    {
        printf("Here:1\n");
        printf("%c\n", h[j]);
        while (n[i] != '\0')
        {
            printf("Here: %d\n", i);
            printf("%c\n", n[i]);
            printf("%c\n", h[j]);
            if (n[i] == h[j])
            {
                printf("Here:2 %d %d\n", i, j);
                printf("%c\n", n[i]);
                printf("%c\n", h[j]);
                b[i + 1] = n[i];
                i = i + 1;
                j = j + 1;

            }
            else
            {
                printf("Here:3\n");
                flag2 = 1;
                break;
            }
        }

        if (flag2 == 1)
        {
            printf("Here:4\n");
            return 0;
        }
        else
        {
            printf("Here:5\n");
            b[i + 1] = '\0';
            s3 = b;
            return 1;
        }

    }
    else
    {
        printf("Here:6\n");
        return 0;
    }

}

Input: s2: vanilla s1: lla

Output: 1

`strcpy` does _not_ allocate memory for you. You should do: `char *h=malloc(strlen(haystack)+1); strcpy(h, haystack);` — Paul Ogilvie, Sep 27 '18 at 08:38
Also your code looks terribly convoluted. And you never use `s3` etc. There are many issues. You should probably start with something simpler and read the chapters dealing with strings and with pointers in your C text book. — Jabberwocky, Sep 27 '18 at 08:45
A `char*` is not some pre-made string class like in other languages. It's just a low level address pointing at memory allocated elsewhere. See https://stackoverflow.com/questions/37549594/crash-or-segmentation-fault-when-data-is-copied-scanned-read-to-an-uninitializ — Lundin, Sep 27 '18 at 08:45
@KrishanPalSingh OK, better now, but please show also how you call your `code` function. Provide a [MCVE]. — Jabberwocky, Sep 27 '18 at 08:57
`char h[] = ""; strcpy(h, s2);` --> `h[]` is only a 1 `char` array. The `stcpy()` will certainly overfill the destination. — chux - Reinstate Monica, Sep 27 '18 at 09:11

score 0 · Answer 1 · answered Sep 27 '18 at 09:29

The char arrays h, n and b hava a size of 1.

Do following modifications in your code:

  char h[100] = "";  // << change here
  strcpy(h, s2);
  printf("%s\n", h);

  char n[100] = "";  // << change here
  strcpy(n, s1);
  printf("%s\n", n);

  char b[100] = "";  // << change here

With these modifications the arrays can contain up to 100 characters. Now your code seems to work more or less.

Disclaimers:

even with these modifications your code is still terrible and inefficient
the modifications are just quick and dirty hacks, the function won't work if the strings involved are longer than 99
there may be other problems in your code
s3 = b; this won't work for several reasons, but that's another topic

How does character array behave when initialized via strcpy?

1 Answers1