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There are two ways to initialize arrays in C: <ul> <li> On the stack (which will handle memory for you since it will be cleaned up when your function ends)</li> <li> In the heap (which will require you to handle allocation and freeing on your own). </li> </ul> If you would like to use the stack, you could initialize your array like this... <pre><code>#define ARRAY_LENGTH 10 void *ptr; void *arr[ARRAY_LENGTH]; for (int i = 0; i < ARRAY_LENGTH; i++) { arr[i] = ptr; } </code></pre> You can similarly define your array in the heap as follows... <pre><code>#define ARRAY_LENGTH 10 void *ptr; void **arr = malloc(sizeof(void *) * ARRAY_LENGTH); for (int i = 0; i < ARRAY_LENGTH; i++) { arr[i] = ptr; } free(arr); </code></pre> It is important to remember that an array (besides arrays assigned in the stack, which have some additional attributes such as length) is essentially just a pointer to the first element, and the operation arr[i] is the same as moving i*sizeof(elem) bytes away from the first element, and accessing the memory there. If you would like to get a pointer to the ith index in the array, then you would use notations such as... <pre><code>void *indexPtr = arr + i; </code></pre> or <pre><code>void *indexPtr = &( arr[i] ); </code></pre> In this fashion, an array of void*'s would be of type void **, since the variable is a pointer to the first member of the array, which is a pointer. This can be a bit confusing, but just always try to keep in mind what type the elements of the array are, and creating a pointer to them. So if the array is of type int, then the array would be of type int or int[], but if you are storing pointers to integers, you would initialize an array of type int * in either of these two forms... <pre><code>int **arr = malloc(sizeof(int *) * ARRAY_LENGTH); int *arr[ARRAY_LENGTH]; </code></pre> Also note that you are storing pointers, so if you run the code... <pre><code>int *arr[4]; for (int i = 0; i < ARRAY_LENGTH; i++) { arr[i] = &i; } </code></pre> Although it may seem to be that the values pointed to in the array would be as follows- [0, 1, 2, 3], but in reality it would be [4, 4, 4, 4], since what you actually have is an array of pointers all pointing to the variable i in your function, so whenever you change that, the values pointed to in the array will all be changed. I hope this helped
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