How should work with arrays of string in c - c

Question

I want to use an array of string the same as this:
char arr[][20] = {"username1", "username2"};
after i have not problem to get values, for example :
printf("%s", arr[0]); // for "username1"
i have problem to insert new string to this array, something like this!? :
arr[2] = "username3"; // or sprintf(arr[2], "%s", "username3");

Answer 1

You're trying to insert a char* into an array of char[20]. That won't work. You could use strcpy as
strcpy(arr[2], "username3");
if you had allocated sufficient space for a third string, which you haven't. I suggest you read up on malloc and friends; you can't just append a string to this array.

Answer 2

With this static initialization, arr will not dynamically resize to hold a new entry (pointers to entries in the stack generally don't like being written to).
In order for this to work, arr should be of char ** and allocated properly:
// allocate enough for three entries in arr
char **arr = (char **) malloc(3 * sizeof(char *));

Answer 3

The wrong way of doing this would be as follows:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void)
{
char arr[][20] = {"username1", "username2"};
printf("%s\n", arr[0]);
printf("%s\n", arr[1]);
printf("sizeof arr : %d \n", sizeof(arr)/sizeof(*arr));
strcpy(arr[2],"username3");
printf("%s\n", arr[2]);
printf("sizeof arr : %d \n", sizeof(arr)/sizeof(*arr));
return 0;
}
because, when you write something like the below
char arr[][20] = {"username1", "username2"};
The first dimension is automatically calculated based on the given initializer, in this case, it is 2 (i.e. index 0 and 1).
Later when you try to access (read or write), the third element i.e. index 2, it is a undefined behavior.
If you want to increase the first dimension of a 2d array, the correct way of doing that would be as follows:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define LEN 20
int main(void)
{
char **arr = NULL;
int n=2, n1;
int i;
if ((arr = malloc(sizeof(char *) * n)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
for (i=0; i<n ; i++) {
if ((arr[i] = malloc(sizeof(char) * LEN)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
}
strcpy(arr[0], "username1");
strcpy(arr[1], "username2");
printf("%s\n", arr[0]);
printf("%s\n", arr[1]);
/* add 5 more elements to arr */
n1 = 5;
if ((arr = realloc(arr, sizeof(char *) * (n + n1))) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
for (i=n; i<n1+n ; i++) {
if ((arr[i] = malloc(sizeof(char) * LEN)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
}
strcpy(arr[2], "username2");
strcpy(arr[3], "username3");
strcpy(arr[4], "username4");
strcpy(arr[5], "username5");
strcpy(arr[6], "username6");
/* if you uncomment the below, it will lead to undefined behavior */
//strcpy(arr[7], "username7");
for (i=0; i<n+n1 ; i++)
printf("%d : %s \n", i, arr[i]);
for (i=0; i<n+n1 ; i++)
free(arr[i]);
free(arr);
return 0;
}
$ ./a.out
username1
username2
0 : username1
1 : username2
2 : username2
3 : username3
4 : username4
5 : username5
6 : username6
$
Hope this helps!

Answer 4

C does not have dynamically growing data structures built-in. If you want an array that can hold an increasing number of elements, you need to involve dynamic memory allocation. This is usually done using the realloc() function, which can be used to allocate and then re-allocate a memory block, so that you can grow the allocation as needed.