Question

I have been trying to execute the following code. Though the task is trivial, I am getting a segmentation fault.

The aim of the code snippet betlow is to create a multi dimensional array of maximum row size 4 and column size 33. Then after creation, it should set the contents of all the rows as 0, followed by a '\0' character. Then in the end, it should display the output on the stdout.

Although I am not new to programming, I keep on getting similar errors, so if possible please explain me how can I avoid such mistakes in the future.

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

int main(int argc, char* argv[])
{
        int i,j,k,x,y;
        char** arr;
        arr = (char**) malloc(4 * sizeof(char*));
        for ( i = 0; i < 4; i++) {
                arr[i] = (char*) malloc(9 * sizeof(char));
                memset(arr,0,8);
                arr[i][8] = '\0';
        }
        for ( j = 0; j<4; j++) {
                puts(arr[j]);
        }
        return 0;
}

Answer 1

You’re memseting the wrong pointer.

Instead of:

memset(arr,0,8);

you want:

memset(arr[i],0,8);

So you’re off by one level of indirection.

---

As pointed out in the comments, here are some optimizations:

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) malloc(9 * sizeof(char));
    memset(arr[i],0,9);
}

or

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) calloc(9, sizeof(char));
}

Note that the cast to char* isn’t necessary in C.

---

If you wanted to the character '0' instead of the null-character, then you should go with this:

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) malloc(8 * sizeof(char));
    memset(arr[i],'0',8);
    arr[i][8] = '\0';
}