memset() was called for array[]. But another array gets emptied in C

Question

I don't understand this thing. about memset in C

I stored john in array[].
Then I created char* megaArray[10] which can store 10 strings.
I assigned megaArray[0] = array. So john gets stored in 0 position of megaArray[].
I called memset() to reset array[]. array gets emptied.
Why does tokens[0] gets emptied if I called memset() for array[] ?

Please Help,

Here is the code

#include <stdio.h>

int main() {
    char array[] = "john";
    printf("%s is the name\n",array);
    
    char* megaArray[10];
    megaArray[0] = array;
    
    printf("%s is the name in mega Array\n",megaArray[0]);
    
    memset(array, '\0', sizeof(array));
    
    printf("%s is the name after memset\n",array);
    
    printf("'%s' is the name in mega Array after memset\n",megaArray[0]);
    
    
    return 0;
}

What do you think, is the difference between `array` and `megaArray[0]`? — Gerhardh, Dec 13 '21 at 10:09
"So john gets stored" No, `john` is not moved or copied anywhere in that instruction. — Gerhardh, Dec 13 '21 at 10:09
I assume `tokens` is a typo and should be `megaArray`, correct? — Gerhardh, Dec 13 '21 at 10:11
You are copying a *pointer*. Remember that an array appearing in most expressions (with some exceptions) gets converted to a pointer to its first element. — Ian Abbott, Dec 13 '21 at 10:11
There is one array containing `"john"`. If you change that, then anything that points there also reports the change. — Weather Vane, Dec 13 '21 at 10:12
See [Common string handling pitfalls in C programming](https://software.codidact.com/posts/284849) for a beginner FAQ. — Lundin, Dec 13 '21 at 10:14

score 2 · Answer 1 · answered Dec 13 '21 at 10:13

The assignment

megaArray[0] = array;

doesn't create a copy of array.

Instead it makes megaArray[0] point to (the first element of) array. It's equivalent to:

megaArray[0] = &array[0];

Somewhat graphically it looks like this (after the assignment):


+--------------+     +----------+----------+-----+----------+
| megaArray[0] | --> | array[0] | array[1] | ... | array[4] |
+--------------+     +----------+----------+-----+----------+
| megaArray[1] | --> ?
+--------------+
| ...          |
+--------------+
| megaArray[9] | --> ?
+--------------+

score 1 · Answer 2 · answered Dec 13 '21 at 10:12

char* megaArray[10];

is an array of char-pointers. So when you assign

megaArray[0] = array;

You let the first array element "point" to the string stored in the variable array. It does not copy the string. megaArray[0] now contains an address to array and its literaly the same string at the same memory location. So memsetting array changes one and the same string.

score 0 · Answer 3 · answered Dec 13 '21 at 10:16

char* megaArray[10]; is simple array of pointers.

megaArray[0] = array; it assigns the pointer megaArray[0] with reference (address) of array. It does not allocate new memory or copy the data.

printf("'%s' is the name in mega Array after memset\n",megaArray[0]);

prints the array which was zeroed by memset as megaArray[0] contains a pointer to array

If you want to make the copy of the string held in array

megaArray[0] = strdup(array);

or

megaArray[0] = malloc(sizeof(array));
memcpy(megaArray[0], array, sizeof(array));

memset() was called for array[]. But another array gets emptied in C

3 Answers3