segfault when setting character of malloced pointer-to-pointer string

Question

In a bigger project i have been working on, i got an Segmentation fault (core dumped), so i managed to reproduce the problem like this:

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

int main() {
    char **x = malloc(10 * 10);
    x[0][0] = 'a';

    printf("%s\n", x[0]);

    free(x);

    return 0;
}

Output: Segmentation fault (core dumped)

For some reason, (i don't know why) i get a Segmentation fault (core dumped) when i set a character in a pointer-to-pointer string (that is malloced)

Since the pointer is allocated on the heap, shouldn't i be able to change the array elements?

So i tried to initalize the array using memset like this:

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

int main() {
    char **x = malloc(10 * 10);
    memset(x[0], 0x00, 10);
    x[0][0] = 'a';

    printf("%s", x[0]);

    free(x);

    return 0;
}

Output: Segmentation fault (core dumped)

Maybe i need to use strcpy?

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

int main() {
    char **x = malloc(10 * 10);
    strcpy(x[0], "Hi");

    printf("%s", x[0]);

    free(x);

    return 0;
}

Output: Segmentation fault (core dumped)

Okay maybe pointer-to-pointer just does not work even if it is on the heap.

What if it is a normal string?

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

int main() {
    char *x = malloc(10);
    x[0] = 'a';

    printf("%s\n", x);

    free(x);

    return 0;
}

Output: a

So why does a normal pointer work??

Maybe the internet will help.

Nevermind after i got bored before finding a single solution on the internet.

Can someone find a solution? Im not a genius at how C works so do forgive me if said something wrong.

MikeCAT · Accepted Answer · 2022-07-23 00:32:06Z

3

In this line

char **x = malloc(10 * 10);

you are allocating an uninitialized array of char*.

x[0][0] is dereferencing an uninitialized element x[0] of the array, so this will result in Segmentation Fault.

You should allocate some buffer and assign that to the array before copying strings.

Also don't forget to add terminating null-character to use the data as strings.

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

int main() {
    char **x = malloc(sizeof(*x) * 10);
    x[0] = malloc(10); /* allocate buffer and assign */
    x[0][0] = 'a';
    x[0][1] = '\0'; /* add terminating null-character */

    printf("%s\n", x[0]);

    free(x[0]);
    free(x);

    return 0;
}

answered Jul 23, 2022 at 0:32

MikeCAT

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1 Comment

iocode Over a year ago

Oh, thanks alot! I did not know that the element dereferenced has to exist first!

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