Can someone give me a quick piece of code that shows me how to properly hand over a struct pointer to a function? I just can't seem to get it right.

Hello,

here is an example.
It sets a struct pointer to an allocated memory area and initializes the struct.

Code:

---

void init_mystruct(MYSTRUCT* mystruct) {
   mystruct.whatever = ...;
}

...

// dynamically allocate memory for a struct and initialize it
MYSTRUCT* mystruct = malloc(sizeof(MYSTRUCT)); 
init_mystruct(mystruct);

---

can anyone tell how to do this but instead of mystruct, do it with a array mystruct[n elements]

You can't pass an array by value, but only by reference.
Code:

---

void f(int* p, int n)
{
   int i = 0;
   for(i = 0; i < n; i++)
   {
      p[ i] = 100; // Write over memory area that is where p points to + sizeof(int) * i. That's where the i'th element is at.
   }
}
int main()
{
   int arr[10];
   f(arr, 10);   // "arr" means: a pointer to the first element.
}

---

Why did you do this part:

Code:

---

MYSTRUCT* mystruct = malloc(sizeof(MYSTRUCT)); 

---

Why did you malloc? Will it not work without malloc?

no, you have to allocate memory when using pointers. look up "pointers" in your c language reference. and the correct way to use structs is with the "struct" keyword before your structure identifier, e.g. struct MYSTRUCT s; or struct MYSTRUCT *s = malloc(sizeof(struct MYSTRUCT)), exception herefore is of course using typedefs for declaring structures.

to allocate an array of ten MYSTRUCTS:
Code:

---

struct MYSTRUCT *structs = malloc(sizeof(struct MYSTRUCT) * 10);

---

now you can pass this array to function f as follows:
Code:

---

void f(struct MYSTRUCT *structs, var count);
//...
f(structs, 10);

---

that is because pointers and arrays have a tight relationship in c. they're basically the same language feature (in plain c, you can replace * by [] nearly at will); the difference in between them is in memory allocation and memory access. declaring and defining var a[3] = {0, 1, 2} gives you the variable a which can't be reassigned but will be converted silently into var *a when passed to a function expecting a pointer as parameter. this conversion will not be done recursively and thus won't work with multidimensional arrays or pointers to pointers.

that's just a brief overview...

joey.