Dynamic array of pointers - arrays

Question

I have an array of pointers to strings.
char **array;
I declare it this way and not char *array[N] because this array won't have a static number of elements.
Declaring the array this way, I will probably have to realloc the sizeof it every time I add a new element (pointer to string).
int main(void)
{
char **array;
char *word = "lolol";
char *word2 = "blabla";
return 0;
}
Could you give me an example on how I should "create space" in the array in order to store pointers to these strings?

Answer 1

The best way of doing it is probably by making a struct
This way, you can resize it, and add as many strings as you want without needing to choose a specific size.
Note: setting the string_array's capacity and size to 0 is necessary for it to work.
You could do it by a function like this instead:
void load_array(string_array *array)
{
array->size = 0;
array->capacity = 0;
}
And call it like this:
load_array(&my_array);
Note, when getting the value from one of these arrays using [], you must call it like this:
my_array.arr[index]
This is because you must refer to the pointer in the array struct, which is as arr (char **)
I have tested the below, and it works perfectly.
# include <stdio.h>
typedef struct string_array
{
char **arr;
unsigned capacity, size;
} string_array;
void add_to_array(string_array *array, char *str)
{
if(array->capacity == 0)
{
array->arr = (char **)malloc((array->capacity = 3) * sizeof(char *));
array->arr[array->size++] = str;
}
else if(array->capacity == array->size)
{
array->arr = (char **)realloc(array->arr, (array->capacity *= 1.5) * sizeof(char *));
array->arr[array->size++] = str;
}
else
{
array->arr[array->size++] = str;
}
}
int main(void)
{
char *str1 = "Hello World";
char *str2 = "Hello World2";
char *str3 = "Hello World3";
char *str4 = "Hello World4";
char *str5 = "Hello World5";
string_array my_array;
my_array.capacity = 0;
my_array.size = 0;
add_to_array(&my_array, str1);
add_to_array(&my_array, str2);
add_to_array(&my_array, str3);
add_to_array(&my_array, str4);
add_to_array(&my_array, str5);
// and so on
for (int i = 0; i < my_array.size; ++i)
{
printf(my_array.arr[i]);
printf("\n");
}
free(my_array.arr);
getchar(); // this means pressing enter closes the console
return (0);
}

Answer 2

Here is a demonstrative program
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
size_t N = 2;
char **array = malloc( N * sizeof( char * ) );
if ( !array ) return 1;
char *word = "lolol";
char *word2 = "blabla";
array[0] = word;
array[1] = word2;
char *word3 = "Hello";
++N;
array = realloc( array, N * sizeof( char * ) );
if ( !array ) return 2;
array[2] = word3;
for ( size_t i = 0; i < N; i++ ) puts( array[i] );
free( array );
return 0;
}
The program output is
lolol
blabla
Hello

Answer 3

Simply allocate some room for pointers in your array, and free it when you're done.
char *word = "lolol";
char *word2 = "blabla";
char** array = malloc(2*sizeof(char*));
array[0] = word;
array[1] = word2;
free(array);
You can change that 2*sizeof(char*) to a N*sizeof(char*) if you want more elements.